IB Questionbank

3.

[Maximum mark: 1]

19M.1A.SL.TZ1.3

A sky diver is falling at terminal speed when she opens her parachute. What are the direction of her velocity vector and the direction of her acceleration vector before she reaches the new terminal speed?

[1]

Markscheme

C

30.

[Maximum mark: 1]

19M.1A.SL.TZ1.30

Most power stations rely on a turbine and a generator to produce electrical energy. Which power station works on a different principle?

A. Nuclear

B. Solar

C. Fossil fuel

D. Wind

[1]

Markscheme

B

4.

[Maximum mark: 1]

19M.1A.SL.TZ1.4

A stone is thrown downwards from the edge of a cliff with a speed of 5.0 m s^–1. It hits the ground 2.0 s later. What is the height of the cliff?

A. 20 m

B. 30 m

C. 40 m

D. 50 m

[1]

Markscheme

B

Examiners report

This question was well answered by the majority of candidates and had a high discrimination index.

5.

[Maximum mark: 1]

19M.1A.SL.TZ1.5

A ball is thrown upwards at an angle to the horizontal. Air resistance is negligible. Which statement about the motion of the ball is correct?

A. The acceleration of the ball changes during its flight.

B. The velocity of the ball changes during its flight.

C. The acceleration of the ball is zero at the highest point.

D. The velocity of the ball is zero at the highest point.

[1]

Markscheme

B

Examiners report

Candidate responses were divided between responses B (correct), D, and to a lesser extent, C. Many candidates appeared to focus on vertical velocity only or confused vertical velocity and acceleration values. This question had the highest discrimination index, suggesting that it would be a useful question for class discussion.

3.

[Maximum mark: 1]

19M.1A.SL.TZ2.3

The graph shows the variation of velocity of a body with time along a straight line.

What is correct for this graph?

A. The maximum acceleration is at P.

B. The average acceleration of the body is given by the area enclosed by the graph and time axis.

C. The maximum displacement is at Q.

D. The total displacement of the body is given by the area enclosed by the graph and time axis.

[1]

Markscheme

D

30.

[Maximum mark: 1]

19M.1A.SL.TZ2.30

The orbital radius of the Earth around the Sun is 1.5 times that of Venus. What is the intensity of solar radiation at the orbital radius of Venus?

A. 0.6 kW m^-2

B. 0.9 kW m^-2

C. 2 kW m^-2

D. 3 kW m^-2

[1]

Markscheme

D

Examiners report

This had a low discrimination index at both SL and HL and although the correct answer was the most popular, all options gained high support. Candidates should be reminded that they have a data booklet and become familiar with its contents before the exam.

31.

[Maximum mark: 1]

19M.1A.SL.TZ2.31

A proton of velocity v enters a region of electric and magnetic fields. The proton is not deflected. An electron and an alpha particle enter the same region with velocity v. Which is correct about the paths of the electron and the alpha particle?

[1]

Markscheme

D

1.

[Maximum mark: 14]

19M.2.SL.TZ2.1

A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of 64 m s^–1. The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms.

The student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.

The following data are available.

Height of P = 2.80 m

Distance of student from net = 11.9 m

Height of net = 0.910 m

Initial speed of tennis ball = 64 m s^-1

(ai)

Calculate the average force exerted by the racquet on the ball.

[2]

Markscheme

$F = \frac{{\Delta mv}}{{\Delta t}}/m\frac{{\Delta v}}{{\Delta t}}/\frac{{0.058 \times 64.0}}{{25 \times {{10}^{ - 3}}}}$ ✔

$F$ = 148« ${\text{N}}$ »≈150« ${\text{N}}$ » ✔

Examiners report

At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.

(aii)

Calculate the average power delivered to the ball during the impact.

[2]

Markscheme

ALTERNATIVE 1

$P = \frac{{\frac{1}{2}m{v^2}}}{t}/\frac{{\frac{1}{2} \times 0.058 \times {{64.0}^2}}}{{25 \times {{10}^{ - 3}}}}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

ALTERNATIVE 2

$P = {\text{average}}Fv/148 \times \frac{{64.0}}{2}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

Examiners report

This was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.

(bi)

Calculate the time it takes the tennis ball to reach the net.

[2]

Markscheme

horizontal component of velocity is 64.0 × cos7° = 63.52 «ms⁻¹» ✔

$t = « \frac{{11.9}}{{63.52}} =» 0.187/0.19 « {\text{s}}$ » ✔

Do not award BCA. Check working.

Do not award ECF from using 64 m s^-1.

Examiners report

This question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.

(bii)

Show that the tennis ball passes over the net.

[3]

Markscheme

ALTERNATIVE 1

u_y= 64 sin7/7.80 «ms⁻¹»✔

decrease in height = 7.80 × 0.187 + $\frac{1}{2}$ × 9.81 × 0.187²/1.63 «m» ✔

final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔

time to fall this distance found using «=1.89 = 7.8t + $\frac{1}{2}$ × 9.81 ×t²»

t = 0.21 «s»✔

0.21 «s» > 0.187 «s» ✔

«reaches the net before it has fallen far enough so goes over»

Other alternatives are possible

Examiners report

There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.

A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.

(biii)

Determine the speed of the tennis ball as it strikes the ground.

[2]

Markscheme

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}$ × 0.058 × 64² + 0.058 × 9.81 × 2.80 = $\frac{1}{2}$ × 0.058 × v² ✔

v = 64.4 «ms⁻¹» ✔

ALTERNATIVE 2

${v_v} = « \sqrt {{{7.8}^2} + 2 \times 9.81 \times 2.8} » = 10.8 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

« $v = \sqrt {{{63.5}^2} + {{10.8}^2}}$ »

$v = 64.4 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

Examiners report

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

(c)

The student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.

The model assumes

• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.

Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.

[3]

Markscheme

so horizontal velocity component at lift off for clay is smaller ✔

normal force is the same so vertical component of velocity is the same ✔

so bounce angle on clay is greater ✔

Examiners report

As the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.

a.

[Maximum mark: 1]

19M.2.SL.TZ2.14a

(a)

Show that, when the speed of the train is 10 m s-1, the frequency of the periodic force is 0.4 Hz.

[1]

Markscheme

time period

T = « $\frac{25}{10}$ » = 2.5 s AND f = $\frac{1}{T}$

OR

evidence of f = $\frac{10}{25}$ ✔

Answer 0.4 Hz is given, check correct working is shown.

Examiners report

The question was correctly answered by almost all candidates.

b.

[Maximum mark: 2]

19M.2.SL.TZ2.14b

(b)

Outline, with reference to the curve, why it is unsafe to drive a train across the bridge at 30 m s^-1 for this amount of damping.

[2]

Markscheme

30 m s^–1 corresponds to f = 1.2 Hz ✔

the amplitude of vibration is a maximum for this speed

OR

corresponds to the resonant frequency ✔

Examiners report

The answers to this question were generally well presented and a correct argument was presented by almost all candidates. Resonance was often correctly referred to.

c.

[Maximum mark: 2]

19M.2.SL.TZ2.14c

(c)

The damping of the bridge system can be varied. Draw, on the graph, a second curve when the damping is larger.

[2]

Markscheme

similar shape with lower amplitude ✔

maximum shifted slightly to left of the original curve ✔

Amplitude must be lower than the original, but allow the amplitude to be equal at the extremes.

Examiners report

A correct curve, with lower amplitude and shifted left, was drawn by most candidates.

a.

[Maximum mark: 1]

19M.2.SL.TZ2.15a

(a)

Identify, on the HR diagram, the position of the Sun. Label the position S.

[1]

Markscheme

the letter S should be in the region of the shaded area ✔

Examiners report

Locating the Sun’s position on the HR diagram was correctly done by most candidates, although a few were unsure of the surface temperature of the Sun.

d.

[Maximum mark: 2]

19M.2.SL.TZ2.15d

(d)

During its evolution, the Sun is likely to be a red giant of surface temperature 3000 K and luminosity 10⁴ L_☉. Later it is likely to be a white dwarf of surface temperature 10 000 K and luminosity 10-4 L_☉. Calculate the $\frac{{{\text{radius of the Sun as a white dwarf}}}}{{{\text{radius of the Sun as a red giant}}}}$ .

[2]

Markscheme

Examiners report

Calculating the ratio of the radius of a white dwarf to a red giant star was done quite well by most candidates. However quite a few candidates made POT errors or forgot to take the final square root.

ai.

[Maximum mark: 2]

19M.2.SL.TZ2.1ai

(ai)

Calculate the average force exerted by the racquet on the ball.

[2]

Markscheme

$F = \frac{{\Delta mv}}{{\Delta t}}/m\frac{{\Delta v}}{{\Delta t}}/\frac{{0.058 \times 64.0}}{{25 \times {{10}^{ - 3}}}}$ ✔

$F$ = 148« ${\text{N}}$ »≈150« ${\text{N}}$ » ✔

Examiners report

At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.

aii.

[Maximum mark: 2]

19M.2.SL.TZ2.1aii

(aii)

Calculate the average power delivered to the ball during the impact.

[2]

Markscheme

ALTERNATIVE 1

$P = \frac{{\frac{1}{2}m{v^2}}}{t}/\frac{{\frac{1}{2} \times 0.058 \times {{64.0}^2}}}{{25 \times {{10}^{ - 3}}}}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

ALTERNATIVE 2

$P = {\text{average}}Fv/148 \times \frac{{64.0}}{2}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

Examiners report

This was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.

bi.

[Maximum mark: 2]

19M.2.SL.TZ2.1bi

(bi)

Calculate the time it takes the tennis ball to reach the net.

[2]

Markscheme

horizontal component of velocity is 64.0 × cos7° = 63.52 «ms⁻¹» ✔

$t = « \frac{{11.9}}{{63.52}} =» 0.187/0.19 « {\text{s}}$ » ✔

Do not award BCA. Check working.

Do not award ECF from using 64 m s^-1.

Examiners report

This question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.

bii.

[Maximum mark: 3]

19M.2.SL.TZ2.1bii

(bii)

Show that the tennis ball passes over the net.

[3]

Markscheme

ALTERNATIVE 1

u_y= 64 sin7/7.80 «ms⁻¹»✔

decrease in height = 7.80 × 0.187 + $\frac{1}{2}$ × 9.81 × 0.187²/1.63 «m» ✔

final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔

time to fall this distance found using «=1.89 = 7.8t + $\frac{1}{2}$ × 9.81 ×t²»

t = 0.21 «s»✔

0.21 «s» > 0.187 «s» ✔

«reaches the net before it has fallen far enough so goes over»

Other alternatives are possible

Examiners report

There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.

A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.

biii.

[Maximum mark: 2]

19M.2.SL.TZ2.1biii

(biii)

Determine the speed of the tennis ball as it strikes the ground.

[2]

Markscheme

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}$ × 0.058 × 64² + 0.058 × 9.81 × 2.80 = $\frac{1}{2}$ × 0.058 × v² ✔

v = 64.4 «ms⁻¹» ✔

ALTERNATIVE 2

${v_v} = « \sqrt {{{7.8}^2} + 2 \times 9.81 \times 2.8} » = 10.8 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

« $v = \sqrt {{{63.5}^2} + {{10.8}^2}}$ »

$v = 64.4 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

Examiners report

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

c.

[Maximum mark: 3]

19M.2.SL.TZ2.1c

(c)

The student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.

The model assumes

• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.

Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.

[3]

Markscheme

so horizontal velocity component at lift off for clay is smaller ✔

normal force is the same so vertical component of velocity is the same ✔

so bounce angle on clay is greater ✔

Examiners report

As the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.

2.

[Maximum mark: 8]

19M.2.SL.TZ2.2

A container of volume 3.2 × 10-6 m³ is filled with helium gas at a pressure of 5.1 × 10⁵ Pa and temperature 320 K. Assume that this sample of helium gas behaves as an ideal gas.

A helium atom has a volume of 4.9 × 10^-31 m³.

(a)

The molar mass of helium is 4.0 g mol^-1. Show that the mass of a helium atom is 6.6 × 10^-27 kg.

[1]

Markscheme

$m = \frac{{4.0 \times {{10}^{ - 3}}}}{{6.02 \times {{10}^{23}}}}$ «kg»

OR

6.64 × 10⁻²⁷«kg» ✔

Examiners report

The mark was awarded for a clear substitution or an answer to at least 3sf. Many gained the mark for a clear substitution with a conversion from g to kg somewhere in their response. Fewer gave the answer to the correct number of sf.

(b)

Estimate the average speed of the helium atoms in the container.

[2]

Markscheme

$\frac{1}{2}m{v^2} = \frac{3}{2}kT/v = \sqrt {\frac{{3kT}}{m}} /\sqrt {\frac{{3 \times 1.38 \times {{10}^{ - 23}} \times 320}}{{6.6 \times {{10}^{ - 27}}}}}$ ✔

v = 1.4 × 10³ «ms⁻¹» ✔

Examiners report

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

(c)

Show that the number of helium atoms in the container is about 4 × 10²⁰.

[2]

Markscheme

$N = \frac{{pV}}{{kT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}}}}{{1.38 \times {{10}^{ - 23}} \times 320}}$

OR

$N = \frac{{pV{N_A}}}{{RT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}} \times 6.02 \times {{10}^{23}}}}{{8.31 \times 320}}$ ✔

N = 3.7 × 10²⁰ ✔

Examiners report

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by N_A to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by N_A.

(di)

Calculate the ratio $\frac{{{\text{total volume of helium atoms}}}}{{{\text{volume of helium gas}}}}$ .

[1]

Markscheme

« $\frac{{4 \times {{10}^{20}} \times 4.9 \times {{10}^{ - 31}}}}{{3.2 \times {{10}^{ - 6}}}} =» 6 \times {10^{ - 5}}$ ✔

Examiners report

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

(dii)

Explain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.

[2]

Markscheme

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

Examiners report

In general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

a.

[Maximum mark: 2]

19M.2.SL.TZ2.2a

(a)

The mass of a helium atom is 6.6 × 10^-27 kg. Estimate the average speed of the helium atoms in the container.

[2]

Markscheme

$\frac{{\text{1}}}{2}m{v^2} = \frac{3}{2}kT/v = \sqrt {\frac{{3kT}}{m}} /\sqrt {\frac{{3 \times 1.38 \times {{10}^{ - 23}} \times 320}}{{6.6 \times {{10}^{ - 27}}}}}$ ✔

v = 1.4 × 10³«ms^–1»^✔

Examiners report

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

b.

[Maximum mark: 2]

19M.2.SL.TZ2.2b

(b)

Estimate the average speed of the helium atoms in the container.

[2]

Markscheme

$\frac{1}{2}m{v^2} = \frac{3}{2}kT/v = \sqrt {\frac{{3kT}}{m}} /\sqrt {\frac{{3 \times 1.38 \times {{10}^{ - 23}} \times 320}}{{6.6 \times {{10}^{ - 27}}}}}$ ✔

v = 1.4 × 10³ «ms⁻¹» ✔

Examiners report

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

c.

[Maximum mark: 2]

19M.2.SL.TZ2.2c

(c)

Show that the number of helium atoms in the container is about 4 × 10²⁰.

[2]

Markscheme

$N = \frac{{pV}}{{kT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}}}}{{1.38 \times {{10}^{ - 23}} \times 320}}$

OR

$N = \frac{{pV{N_A}}}{{RT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}} \times 6.02 \times {{10}^{23}}}}{{8.31 \times 320}}$ ✔

N = 3.7 × 10²⁰ ✔

Examiners report

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by N_A to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by N_A.

ci.

[Maximum mark: 1]

19M.2.SL.TZ2.2ci

(ci)

Calculate the ratio $\frac{{{\text{volume of helium atoms}}}}{{{\text{volume of helium gas}}}}$ .

[1]

Markscheme

« $\frac{{4 \times {{10}^{20}} \times 4.9 \times {{10}^{ - 31}}}}{{3.2 \times {{10}^{ - 6}}}} =$ » $6 \times {10^{ - 5}}$ ✔

Examiners report

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

cii.

[Maximum mark: 2]

19M.2.SL.TZ2.2cii

(cii)

Discuss, by reference to the kinetic model of an ideal gas and the answer to (c)(i), whether the assumption that helium behaves as an ideal gas is justified.

[2]

Markscheme

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (c)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

Examiners report

At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

di.

[Maximum mark: 1]

19M.2.SL.TZ2.2di

(di)

Calculate the ratio $\frac{{{\text{total volume of helium atoms}}}}{{{\text{volume of helium gas}}}}$ .

[1]

Markscheme

« $\frac{{4 \times {{10}^{20}} \times 4.9 \times {{10}^{ - 31}}}}{{3.2 \times {{10}^{ - 6}}}} =» 6 \times {10^{ - 5}}$ ✔

Examiners report

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

dii.

[Maximum mark: 2]

19M.2.SL.TZ2.2dii

(dii)

Explain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.

[2]

Markscheme

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

Examiners report

In general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

7.

[Maximum mark: 4]

19M.2.SL.TZ2.7

The average temperature of ocean surface water is 289 K. Oceans behave as black bodies.

(a)

Show that the intensity radiated by the oceans is about 400 W m^-2.

[1]

Markscheme

5.67 × 10⁻⁸ × 289⁴

OR

= 396 «W m⁻²» ✔

«≈ 400 W m⁻²»

Examiners report

This was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.

(b)

Explain why some of this radiation is returned to the oceans from the atmosphere.

[3]

Markscheme

«most of the radiation emitted by the oceans is in the» infrared ✔

«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔

«the gases» reradiate/re-emit ✔

partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔

Examiners report

For many this was a well-rehearsed answer which succinctly scored full marks. For others too many vague terms were used. There was much talk about energy being trapped or reflected and the ozone layer was often included. The word ‘albedo’ was often written down with no indication of what it means and ‘the albedo effect also featured.

a.

[Maximum mark: 1]

19M.2.SL.TZ2.7a

(a)

Show that the intensity radiated by the oceans is about 400 W m^-2.

[1]

Markscheme

5.67 × 10⁻⁸ × 289⁴

OR

= 396 «W m⁻²» ✔

«≈ 400 W m⁻²»

Examiners report

This was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.

b.

[Maximum mark: 3]

19M.2.SL.TZ2.7b

(b)

Explain why some of this radiation is returned to the oceans from the atmosphere.

[3]

Markscheme

«most of the radiation emitted by the oceans is in the» infrared ✔

«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔

«the gases» reradiate/re-emit ✔

partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔

Examiners report

For many this was a well-rehearsed answer which succinctly scored full marks. For others too many vague terms were used. There was much talk about energy being trapped or reflected and the ozone layer was often included. The word ‘albedo’ was often written down with no indication of what it means and ‘the albedo effect also featured.

ci.

[Maximum mark: 2]

19M.2.SL.TZ2.7ci

(ci)

Calculate the additional intensity that must be lost by the oceans so that the water temperature remains constant.

[2]

Markscheme

water loses 396 − 330/66 «W m⁻²» ✔

extra intensity that must be lost is «170 − 66» = 104 ≈ 100 «W m⁻²» ✔

OR

absorbed by water 330 + 170/500 «W m⁻²»✔

extra intensity that must be lost is «500 − 396» = 104 ≈ 100 «W m⁻²» ✔

Examiners report

This was well-answered, a very straightforward 2 marks.

cii.

[Maximum mark: 1]

19M.2.SL.TZ2.7cii

(cii)

Suggest a mechanism by which the additional intensity can be lost.

[1]

Markscheme

conduction to the air above

OR

«mainly» evaporation

OR

melting ice at the poles

OR

reflection of sunlight off the surface of the ocean ✔

Do not accept convection or radiation.

Examiners report

Many candidates didn’t understand this question and thought that the answer needed to be some form of human activity that would reduce global temperature rise.

b.

[Maximum mark: 2]

19M.2.SL.TZ2.9b

(b)

The diagram shows some of the electric field lines for two fixed, charged particles X and Y.

The magnitude of the charge on X is $Q$ and that on Y is $q$ . The distance between X and Y is 0.600 m. The distance between P and Y is 0.820 m.

At P the electric field is zero. Determine, to one significant figure, the ratio $\frac{Q}{q}$ .

[2]

Markscheme

$\frac{{kQ}}{{{{(0.600 + 0.820)}^2}}} = \frac{{kq}}{{{{0.820}^2}}}$ ✔

$\frac{Q}{q} = « \frac{{{{(0.600 + 0.820)}^2}}}{{{{0.820}^2}}} = 2.9988 \approx » 3$ ✔

Examiners report

The majority of candidates had an idea of the basic technique here but it was surprisingly common to see the squared missing from the expression for field strengths.

a.

[Maximum mark: 1]

19M.2.SL.TZ2.a

Show that the intensity radiated by the oceans is about 400 W m^-2.

[1]

Markscheme

5.67 × 10⁻⁸ × 289⁴

OR

= 396 «W m⁻²» ✔

«≈ 400 W m⁻²»

Examiners report

This was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.

ai.

[Maximum mark: 2]

19M.2.SL.TZ2.ai

Calculate the average force exerted by the racquet on the ball.

[2]

Markscheme

$F = \frac{{\Delta mv}}{{\Delta t}}/m\frac{{\Delta v}}{{\Delta t}}/\frac{{0.058 \times 64.0}}{{25 \times {{10}^{ - 3}}}}$ ✔

$F$ = 148« ${\text{N}}$ »≈150« ${\text{N}}$ » ✔

Examiners report

At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.

aii.

[Maximum mark: 2]

19M.2.SL.TZ2.aii

Calculate the average power delivered to the ball during the impact.

[2]

Markscheme

ALTERNATIVE 1

$P = \frac{{\frac{1}{2}m{v^2}}}{t}/\frac{{\frac{1}{2} \times 0.058 \times {{64.0}^2}}}{{25 \times {{10}^{ - 3}}}}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

ALTERNATIVE 2

$P = {\text{average}}Fv/148 \times \frac{{64.0}}{2}$ ✔

$P = 4700/4800«{\text{W}}$ » ✔

Examiners report

This was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.

b.

[Maximum mark: 2]

19M.2.SL.TZ2.b

The diagram shows some of the electric field lines for two fixed, charged particles X and Y.

The magnitude of the charge on X is $Q$ and that on Y is $q$ . The distance between X and Y is 0.600 m. The distance between P and Y is 0.820 m.

At P the electric field is zero. Determine, to one significant figure, the ratio $\frac{Q}{q}$ .

[2]

Markscheme

$\frac{{kQ}}{{{{(0.600 + 0.820)}^2}}} = \frac{{kq}}{{{{0.820}^2}}}$ ✔

$\frac{Q}{q} = « \frac{{{{(0.600 + 0.820)}^2}}}{{{{0.820}^2}}} = 2.9988 \approx » 3$ ✔

Examiners report

The majority of candidates had an idea of the basic technique here but it was surprisingly common to see the squared missing from the expression for field strengths.

bi.

[Maximum mark: 2]

19M.2.SL.TZ2.bi

Calculate the time it takes the tennis ball to reach the net.

[2]

Markscheme

horizontal component of velocity is 64.0 × cos7° = 63.52 «ms⁻¹» ✔

$t = « \frac{{11.9}}{{63.52}} =» 0.187/0.19 « {\text{s}}$ » ✔

Do not award BCA. Check working.

Do not award ECF from using 64 m s^-1.

Examiners report

This question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.

bii.

[Maximum mark: 3]

19M.2.SL.TZ2.bii

Show that the tennis ball passes over the net.

[3]

Markscheme

ALTERNATIVE 1

u_y= 64 sin7/7.80 «ms⁻¹»✔

decrease in height = 7.80 × 0.187 + $\frac{1}{2}$ × 9.81 × 0.187²/1.63 «m» ✔

final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔

time to fall this distance found using «=1.89 = 7.8t + $\frac{1}{2}$ × 9.81 ×t²»

t = 0.21 «s»✔

0.21 «s» > 0.187 «s» ✔

«reaches the net before it has fallen far enough so goes over»

Other alternatives are possible

Examiners report

There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.

A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.

biii.

[Maximum mark: 2]

19M.2.SL.TZ2.biii

Determine the speed of the tennis ball as it strikes the ground.

[2]

Markscheme

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}$ × 0.058 × 64² + 0.058 × 9.81 × 2.80 = $\frac{1}{2}$ × 0.058 × v² ✔

v = 64.4 «ms⁻¹» ✔

ALTERNATIVE 2

${v_v} = « \sqrt {{{7.8}^2} + 2 \times 9.81 \times 2.8} » = 10.8 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

« $v = \sqrt {{{63.5}^2} + {{10.8}^2}}$ »

$v = 64.4 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

Examiners report

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

c.

[Maximum mark: 2]

19M.2.SL.TZ2.c

Show that the number of helium atoms in the container is about 4 × 10²⁰.

[2]

Markscheme

$N = \frac{{pV}}{{kT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}}}}{{1.38 \times {{10}^{ - 23}} \times 320}}$

OR

$N = \frac{{pV{N_A}}}{{RT}}/\frac{{5.1 \times {{10}^5} \times 3.2 \times {{10}^{ - 6}} \times 6.02 \times {{10}^{23}}}}{{8.31 \times 320}}$ ✔

N = 3.7 × 10²⁰ ✔

Examiners report

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by N_A to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by N_A.

ci.

[Maximum mark: 2]

19M.2.SL.TZ2.ci

Calculate the additional intensity that must be lost by the oceans so that the water temperature remains constant.

[2]

Markscheme

water loses 396 − 330/66 «W m⁻²» ✔

extra intensity that must be lost is «170 − 66» = 104 ≈ 100 «W m⁻²» ✔

OR

absorbed by water 330 + 170/500 «W m⁻²»✔

extra intensity that must be lost is «500 − 396» = 104 ≈ 100 «W m⁻²» ✔

Examiners report

This was well-answered, a very straightforward 2 marks.

cii.

[Maximum mark: 1]

19M.2.SL.TZ2.cii

Suggest a mechanism by which the additional intensity can be lost.

[1]

Markscheme

conduction to the air above

OR

«mainly» evaporation

OR

melting ice at the poles

OR

reflection of sunlight off the surface of the ocean ✔

Do not accept convection or radiation.

Examiners report

Many candidates didn’t understand this question and thought that the answer needed to be some form of human activity that would reduce global temperature rise.

d.

[Maximum mark: 2]

19M.2.SL.TZ2.d

During its evolution, the Sun is likely to be a red giant of surface temperature 3000 K and luminosity 10⁴ L_☉. Later it is likely to be a white dwarf of surface temperature 10 000 K and luminosity 10-4 L_☉. Calculate the $\frac{{{\text{radius of the Sun as a white dwarf}}}}{{{\text{radius of the Sun as a red giant}}}}$ .

[2]

Markscheme

Examiners report

Calculating the ratio of the radius of a white dwarf to a red giant star was done quite well by most candidates. However quite a few candidates made POT errors or forgot to take the final square root.

di.

[Maximum mark: 1]

19M.2.SL.TZ2.di

Calculate the ratio $\frac{{{\text{total volume of helium atoms}}}}{{{\text{volume of helium gas}}}}$ .

[1]

Markscheme

« $\frac{{4 \times {{10}^{20}} \times 4.9 \times {{10}^{ - 31}}}}{{3.2 \times {{10}^{ - 6}}}} =» 6 \times {10^{ - 5}}$ ✔

Examiners report

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

dii.

[Maximum mark: 2]

19M.2.SL.TZ2.dii

Explain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.

[2]

Markscheme

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

Examiners report

In general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

3.

[Maximum mark: 1]

19N.1A.SL.TZ0.3

A ball falls from rest in the absence of air resistance. The position of the centre of the ball is determined at one-second intervals from the instant at which it is released. What are the distances, in metres, travelled by the centre of the ball during each second for the first 4.0 s of the motion?

A. 5, 10, 15, 20

B. 5, 15, 25, 35

C. 5, 20, 45, 80

D. 5, 25, 70, 150

[1]

Markscheme

B

30.

[Maximum mark: 1]

19N.1A.SL.TZ0.30

What is meant by the statement that the average albedo of the Moon is 0.1?

A. 10% of the radiation incident on the Moon is absorbed by its surface

B. 10% of the radiation emitted by the Moon is absorbed by its atmosphere

C. 10% of the radiation incident on the Moon is reflected by its surface

D. 10% of the radiation emitted by the Moon is at infrared wavelengths

[1]

Markscheme

C

31.

[Maximum mark: 1]

19N.1A.SL.TZ0.31

The force acting between two point charges is $F$ when the separation of the charges is $x$ . What is the force between the charges when the separation is increased to $3 x$ ?

A. $\frac{F}{3}$

B. $\frac{F}{3 x^{2}}$

C. $\frac{F}{9}$

D. $\frac{F}{9 x^{2}}$

[1]

Markscheme

C

32.

[Maximum mark: 1]

19N.1A.SL.TZ0.32

An electron enters a uniform electric field of strength E with a velocity v. The direction of v is not parallel to E. What is the path of the electron after entering the field?

A. Circular

B. Parabolic

C. Parallel to E

D. Parallel to v

[1]

Markscheme

B

4.

[Maximum mark: 1]

19N.1A.SL.TZ0.4

The variation with time t of the acceleration a of an object is shown.

What is the change in velocity of the object from t = 0 to t = 6 s?

A. 6 m s^–1

B. 8 m s^–1

C. 10 m s^–1

D. 14 m s^–1

[1]

Markscheme

C

1.

[Maximum mark: 9]

19N.2.SL.TZ0.1

The graph shows the variation with time t of the horizontal force F exerted on a tennis ball by a racket.

The tennis ball was stationary at the instant when it was hit. The mass of the tennis ball is 5.8 × 10^–2 kg. The area under the curve is 0.84 N s.

(a)

Calculate the speed of the ball as it leaves the racket.

[2]

Markscheme

links 0.84 to Δp ✔

$v =$ « $\frac{0.84}{5.8 \times 10^{- 2}} =$ » 14.5 «m s^–1»✔

NOTE: Award [2] for bald correct answer

(b)

Show that the average force exerted on the ball by the racket is about 50 N.

[2]

Markscheme

use of Δt = «(28 – 12) × 10^–3=» 16 × 10^–3 «s» ✔

$\bar{F}$ =« $\frac{∆ p}{∆ t} =$ » $\frac{0.84}{16 \times 10^{- 3}}$ OR 53 «N» ✔

NOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval

(c)

Determine, with reference to the work done by the average force, the horizontal distance travelled by the ball while it was in contact with the racket.

[3]

Markscheme

E_k= $\frac{1}{2}$ × 5.8 × 10^–2× 14.5²✔

E_k = W ✔

s = « $\frac{W}{F} = \frac{\frac{1}{2} \times 5.8 \times 10^{- 2} \times 14 . 5^{2}}{53} =$ » 0.12 « m » ✔

Allow ECF from (a) and (b)

Allow ECF from MP1

Award [2] max for a calculation without reference to work done, eg: average velocity × time

(d)

Draw a graph to show the variation with t of the horizontal speed v of the ball while it was in contact with the racket. Numbers are not required on the axes.

[2]

Markscheme

graph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔

a.

[Maximum mark: 2]

19N.2.SL.TZ0.1a

(a)

Calculate the speed of the ball as it leaves the racket.

[2]

Markscheme

links 0.84 to Δp ✔

$v =$ « $\frac{0.84}{5.8 \times 10^{- 2}} =$ » 14.5 «m s^–1»✔

NOTE: Award [2] for bald correct answer

b.

[Maximum mark: 2]

19N.2.SL.TZ0.1b

(b)

Show that the average force exerted on the ball by the racket is about 50 N.

[2]

Markscheme

use of Δt = «(28 – 12) × 10^–3=» 16 × 10^–3 «s» ✔

$\bar{F}$ =« $\frac{∆ p}{∆ t} =$ » $\frac{0.84}{16 \times 10^{- 3}}$ OR 53 «N» ✔

NOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval

c.

[Maximum mark: 3]

19N.2.SL.TZ0.1c

(c)

Determine, with reference to the work done by the average force, the horizontal distance travelled by the ball while it was in contact with the racket.

[3]

Markscheme

E_k= $\frac{1}{2}$ × 5.8 × 10^–2× 14.5²✔

E_k = W ✔

s = « $\frac{W}{F} = \frac{\frac{1}{2} \times 5.8 \times 10^{- 2} \times 14 . 5^{2}}{53} =$ » 0.12 « m » ✔

Allow ECF from (a) and (b)

Allow ECF from MP1

Award [2] max for a calculation without reference to work done, eg: average velocity × time

d.

[Maximum mark: 2]

19N.2.SL.TZ0.1d

(d)

Draw a graph to show the variation with t of the horizontal speed v of the ball while it was in contact with the racket. Numbers are not required on the axes.

[2]

Markscheme

graph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔

3.

[Maximum mark: 7]

19N.2.SL.TZ0.3

The solid line in the graph shows the variation with distance $x$ of the displacement $y$ of a travelling wave at t = 0. The dotted line shows the wave 0.20 ms later. The period of the wave is longer than 0.20 ms.

One end of a string is attached to an oscillator and the other is fixed to a wall. When the frequency of the oscillator is 360 Hz the standing wave shown is formed on the string.

Point X (not shown) is a point on the string at a distance of 10 cm from the oscillator.

(a(i))

Calculate, in m s^–1, the speed for this wave.

[1]

Markscheme

v = « $\frac{0.05}{0.20 \times 10^{- 3}} =$ » 250 «m s^–1»✔

(a(ii))

Calculate, in Hz, the frequency for this wave.

[2]

Markscheme

λ = 0.30 «m» ✔
$f$ = « $\frac{250}{0.30} =$ » 830 «Hz» ✔

NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2

(b)

The graph also shows the displacement of two particles, P and Q, in the medium at t = 0. State and explain which particle has the larger magnitude of acceleration at t = 0.

[2]

Markscheme

Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔

(c(i))

State the number of all other points on the string that have the same amplitude and phase as X.

[1]

Markscheme

3 «points» ✔

(c(ii))

The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.

[1]

Markscheme

first harmonic mode drawn ✔

NOTE: Allow if only one curve drawn, either solid or dashed.

a(i).

[Maximum mark: 1]

19N.2.SL.TZ0.3a(i)

(a(i))

Calculate, in m s^–1, the speed for this wave.

[1]

Markscheme

v = « $\frac{0.05}{0.20 \times 10^{- 3}} =$ » 250 «m s^–1»✔

a(ii).

[Maximum mark: 2]

19N.2.SL.TZ0.3a(ii)

(a(ii))

Calculate, in Hz, the frequency for this wave.

[2]

Markscheme

λ = 0.30 «m» ✔
$f$ = « $\frac{250}{0.30} =$ » 830 «Hz» ✔

NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2

b.

[Maximum mark: 2]

19N.2.SL.TZ0.3b

(b)

The graph also shows the displacement of two particles, P and Q, in the medium at t = 0. State and explain which particle has the larger magnitude of acceleration at t = 0.

[2]

Markscheme

Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔

c(i).

[Maximum mark: 1]

19N.2.SL.TZ0.3c(i)

(c(i))

State the number of all other points on the string that have the same amplitude and phase as X.

[1]

Markscheme

3 «points» ✔

c(ii).

[Maximum mark: 1]

19N.2.SL.TZ0.3c(ii)

(c(ii))

The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.

[1]

Markscheme

first harmonic mode drawn ✔

NOTE: Allow if only one curve drawn, either solid or dashed.

4.

[Maximum mark: 8]

19N.2.SL.TZ0.4

A proton is moving in a region of uniform magnetic field. The magnetic field is directed into the plane of the paper. The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton.

The speed of the proton is 2.0 × 10⁶ m s^–1 and the magnetic field strength B is 0.35 T.

(a)

Explain why the path of the proton is a circle.

[2]

Markscheme

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

force is centripetal/towards the centre ✔

NOTE: Accept reference to acceleration instead of force

(b(i))

Show that the radius of the path is about 6 cm.

[2]

Markscheme

$q v B = \frac{m v^{2}}{R}$ ✔

$R = \frac{1.67 \times 10^{- 27} \times 2.0 \times 10^{6}}{1.6 \times 10^{- 19} \times 0.35}$ OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

(b(ii))

Calculate the time for one complete revolution.

[2]

Markscheme

$T = \frac{2 π R}{v}$ ✔

$T = « \frac{2 π \times 0.06}{2.0 \times 10^{6}} = 1.9 \times 10^{- 7}$ « s » ✔

NOTE: Award [2] for bald correct answer

(c)

Explain why the kinetic energy of the proton is constant.

[2]

Markscheme

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence E_k remains constant

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

a.

[Maximum mark: 2]

19N.2.SL.TZ0.4a

(a)

Explain why the path of the proton is a circle.

[2]

Markscheme

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

force is centripetal/towards the centre ✔

NOTE: Accept reference to acceleration instead of force

b(i).

[Maximum mark: 2]

19N.2.SL.TZ0.4b(i)

(b(i))

Show that the radius of the path is about 6 cm.

[2]

Markscheme

$q v B = \frac{m v^{2}}{R}$ ✔

$R = \frac{1.67 \times 10^{- 27} \times 2.0 \times 10^{6}}{1.6 \times 10^{- 19} \times 0.35}$ OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

b(ii).

[Maximum mark: 2]

19N.2.SL.TZ0.4b(ii)

(b(ii))

Calculate the time for one complete revolution.

[2]

Markscheme

$T = \frac{2 π R}{v}$ ✔

$T = « \frac{2 π \times 0.06}{2.0 \times 10^{6}} = 1.9 \times 10^{- 7}$ « s » ✔

NOTE: Award [2] for bald correct answer

c.

[Maximum mark: 2]

19N.2.SL.TZ0.4c

(c)

Explain why the kinetic energy of the proton is constant.

[2]

Markscheme

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence E_k remains constant

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

5.

[Maximum mark: 7]

19N.2.SL.TZ0.5

An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.

(a)

Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10⁸ N C^–1.

[2]

Markscheme

$E = \frac{k \times q}{r^{2}}$ ✔

$E = \frac{8.99 \times 10^{9} \times 6.0 \times 10^{- 3}}{0 . 4^{2}}$ OR $E = 3.37 \times 10^{8} « N C^{- 1} »$ ✔

NOTE: Ignore any negative sign.

(b(i))

Calculate the magnitude of the initial acceleration of the electron.

[2]

Markscheme

$F = q \times E$ OR $F = 1.6 \times 10^{- 19} \times 3.4 \times 10^{8} = 5.4 \times 10^{- 11} « N »$ ✔

$a = « \frac{5.4 \times 10^{- 11}}{9.1 \times 10^{- 31}} = » 5.9 \times 10^{19} « m s^{- 2} »$ ✔

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to $a = « m s^{- 2} »$
Award [2] for bald correct answer

(b(ii))

Describe the subsequent motion of the electron.

[3]

Markscheme

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

a.

[Maximum mark: 2]

19N.2.SL.TZ0.5a

(a)

Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10⁸ N C^–1.

[2]

Markscheme

$E = \frac{k \times q}{r^{2}}$ ✔

$E = \frac{8.99 \times 10^{9} \times 6.0 \times 10^{- 3}}{0 . 4^{2}}$ OR $E = 3.37 \times 10^{8} « N C^{- 1} »$ ✔

NOTE: Ignore any negative sign.

b(i).

[Maximum mark: 2]

19N.2.SL.TZ0.5b(i)

(b(i))

Calculate the magnitude of the initial acceleration of the electron.

[2]

Markscheme

$F = q \times E$ OR $F = 1.6 \times 10^{- 19} \times 3.4 \times 10^{8} = 5.4 \times 10^{- 11} « N »$ ✔

$a = « \frac{5.4 \times 10^{- 11}}{9.1 \times 10^{- 31}} = » 5.9 \times 10^{19} « m s^{- 2} »$ ✔

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to $a = « m s^{- 2} »$
Award [2] for bald correct answer

b(ii).

[Maximum mark: 3]

19N.2.SL.TZ0.5b(ii)

(b(ii))

Describe the subsequent motion of the electron.

[3]

Markscheme

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

7.

[Maximum mark: 7]

19N.2.SL.TZ0.7

A stationary nucleus of uranium-238 undergoes alpha decay to form thorium-234.

The following data are available.

Energy released in decay 4.27 MeV
Binding energy per nucleon for helium 7.07 MeV
Binding energy per nucleon for thorium 7.60 MeV

(a)

Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.

Random:

Spontaneous:

[2]

Markscheme

random:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔

NOTE: OWTTE

spontaneous:
the decay cannot be influenced/modified in any way ✔

NOTE: OWTTE

(b(i))

Calculate the binding energy per nucleon for uranium-238.

[3]

Markscheme

234 × 7.6 OR 4 × 7.07 ✔

BE_U=« 234 × 7.6 + 4 × 7.07 – 4.27 =» $1802$ « MeV » ✔

$\frac{B E_{U}}{A} = « \frac{1802}{238} = » 7.57$ « MeV » ✔

NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10^–12

(b(ii))

Calculate the ratio $\frac{kinetic energy of alpha particle}{kinetic energy of thorium nucleus}$ .

[2]

Markscheme

states or applies conservation of momentum ✔

ratio is « $\frac{E_{k α}}{E_{kTh}} = \frac{\frac{p^{2}}{2 m_{α}}}{\frac{p^{2}}{2 m_{Th}}} = \frac{234}{4}$ » 58.5 ✔

NOTE: Award [2] for bald correct answer

a.

[Maximum mark: 2]

19N.2.SL.TZ0.7a

(a)

Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.

Random:

Spontaneous:

[2]

Markscheme

random:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔

NOTE: OWTTE

spontaneous:
the decay cannot be influenced/modified in any way ✔

NOTE: OWTTE

b(i).

[Maximum mark: 3]

19N.2.SL.TZ0.7b(i)

(b(i))

Calculate the binding energy per nucleon for uranium-238.

[3]

Markscheme

234 × 7.6 OR 4 × 7.07 ✔

BE_U=« 234 × 7.6 + 4 × 7.07 – 4.27 =» $1802$ « MeV » ✔

$\frac{B E_{U}}{A} = « \frac{1802}{238} = » 7.57$ « MeV » ✔

NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10^–12

b(ii).

[Maximum mark: 2]

19N.2.SL.TZ0.7b(ii)

(b(ii))

Calculate the ratio $\frac{kinetic energy of alpha particle}{kinetic energy of thorium nucleus}$ .

[2]

Markscheme

states or applies conservation of momentum ✔

ratio is « $\frac{E_{k α}}{E_{kTh}} = \frac{\frac{p^{2}}{2 m_{α}}}{\frac{p^{2}}{2 m_{Th}}} = \frac{234}{4}$ » 58.5 ✔

NOTE: Award [2] for bald correct answer

a.

[Maximum mark: 2]

19N.2.SL.TZ0.a

Calculate the speed of the ball as it leaves the racket.

[2]

Markscheme

links 0.84 to Δp ✔

$v =$ « $\frac{0.84}{5.8 \times 10^{- 2}} =$ » 14.5 «m s^–1»✔

NOTE: Award [2] for bald correct answer

a(i).

[Maximum mark: 1]

19N.2.SL.TZ0.a(i)

Calculate, in m s^–1, the speed for this wave.

[1]

Markscheme

v = « $\frac{0.05}{0.20 \times 10^{- 3}} =$ » 250 «m s^–1»✔

a(ii).

[Maximum mark: 2]

19N.2.SL.TZ0.a(ii)

Calculate, in Hz, the frequency for this wave.

[2]

Markscheme

λ = 0.30 «m» ✔
$f$ = « $\frac{250}{0.30} =$ » 830 «Hz» ✔

NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2

b.

[Maximum mark: 2]

19N.2.SL.TZ0.b

The graph also shows the displacement of two particles, P and Q, in the medium at t = 0. State and explain which particle has the larger magnitude of acceleration at t = 0.

[2]

Markscheme

Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔

b(i).

[Maximum mark: 2]

19N.2.SL.TZ0.b(i)

Show that the radius of the path is about 6 cm.

[2]

Markscheme

$q v B = \frac{m v^{2}}{R}$ ✔

$R = \frac{1.67 \times 10^{- 27} \times 2.0 \times 10^{6}}{1.6 \times 10^{- 19} \times 0.35}$ OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

b(ii).

[Maximum mark: 3]

19N.2.SL.TZ0.b(ii)

Describe the subsequent motion of the electron.

[3]

Markscheme

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

c.

[Maximum mark: 2]

19N.2.SL.TZ0.c

Explain why the kinetic energy of the proton is constant.

[2]

Markscheme

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence E_k remains constant

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

c(i).

[Maximum mark: 1]

19N.2.SL.TZ0.c(i)

State the number of all other points on the string that have the same amplitude and phase as X.

[1]

Markscheme

3 «points» ✔

c(ii).

[Maximum mark: 1]

19N.2.SL.TZ0.c(ii)

The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.

[1]

Markscheme

first harmonic mode drawn ✔

NOTE: Allow if only one curve drawn, either solid or dashed.

d.

[Maximum mark: 2]

19N.2.SL.TZ0.d

Draw a graph to show the variation with t of the horizontal speed v of the ball while it was in contact with the racket. Numbers are not required on the axes.

[2]

Markscheme

graph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔

4.

[Maximum mark: 1]

20N.1A.SL.TZ0.4

An object of mass $1 kg$ is thrown downwards from a height of $20 m$ . The initial speed of the object is $6 m s^{- 1}$ .
The object hits the ground at a speed of $20 m s^{- 1}$ . Assume $g = 10 m s^{- 2}$ . What is the best estimate of the energy transferred from the object to the air as it falls?

A. $6 J$

B. $18 J$

C. $182 J$

D. $200 J$

[1]

Markscheme

B

40.

[Maximum mark: 1]

20N.1A.SL.TZ0.40

The Rutherford-Geiger-Marsden experiment shows that

A. alpha particles do not obey Coulomb’s law.

B. there is a fixed nuclear radius for each nucleus.

C. a large proportion of alpha particles are undeflected.

D. the Bohr model of the hydrogen atom is confirmed.

[1]

Markscheme

C

Examiners report

This has a low discrimination index but it was felt that perhaps as it was the last question students were guessing the answer especially those choosing option A.

6.

[Maximum mark: 1]

20N.1A.SL.TZ0.6

P and Q leave the same point, travelling in the same direction. The graphs show the variation with time $t$ of velocity $v$ for both P and Q.

What is the distance between P and Q when $t = 8.0 s$ ?

A. $20 m$

B. $40 m$

C. $60 m$

D. $120 m$

[1]

Markscheme

B

8.

[Maximum mark: 1]

20N.1A.SL.TZ0.8

A balloon rises at a steady vertical velocity of $10 m s^{- 1}$ . An object is dropped from the balloon at a height of $40 m$ above the ground. Air resistance is negligible. What is the time taken for the object to hit the ground?

A. $10 s$

B. $5 s$

C. $4 s$

D. $2 s$

[1]

Markscheme

C

Examiners report

Even though over half the candidates are choosing the correct response it has a low discrimination index. Many are choosing D indicating that they forgot to take the velocity upward as negative.

1.

[Maximum mark: 8]

20N.2.SL.TZ0.1

A company delivers packages to customers using a small unmanned aircraft. Rotating horizontal blades exert a force on the surrounding air. The air above the aircraft is initially stationary.

The air is propelled vertically downwards with speed $v$ . The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is $0.95 kg$ and the combined mass of the package and string is $0.45 kg$ . The mass of air pushed downwards by the blades in one second is $1.7 kg$ .

(a(i))

State the value of the resultant force on the aircraft when hovering.

[1]

Markscheme

zero ✓

(a(ii))

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

[2]

Markscheme

Blades exert a downward force on the air ✓

air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓

Downward direction required for MP1.

(a(iii))

Determine $v$ . State your answer to an appropriate number of significant figures.

[3]

Markscheme

«lift force/change of momentum in one second» $= 1.7 v$ ✓

$1.7 v = (0.95 + 0.45) \times 9.81$ ✓

$v = 8.1 « {ms}^{- 1} »$ AND answer expressed to $2$ sf only ✓

Allow $8.2$ from $g = 10 m s^{- 2}$ .

(b)

The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.

[2]

Markscheme

vertical force = lift force – weight OR $= 0.45 \times 9.81$ OR $= 4.4 « N »$ ✓

acceleration $= \frac{0.45 \times 9.81}{0.95} = 4.6 « {ms}^{- 2} »$ ✓

b.

[Maximum mark: 3]

20N.2.SL.TZ0.12b

(b)

Determine the terminal velocity of the sphere.

[3]

Markscheme

radius of sphere $= 0.012 « m »$ ✓

weight of sphere $= 6 π η r v + ρ V g$

OR

$v = \frac{(1.26 \times 10^{- 2} - 915 \times 7.24 \times 10^{- 6}) \times 9.81}{6 π \times 37.9 \times 10^{- 3} \times 1.2 \times 10^{- 2}}$ ✓

$v = 6.84 « m s^{- 1} »$ ✓

Accept use of $g = 10$ leading to $v = 7.0 « m s^{- 1} »$

Allow implicit calculation of radius for MP1

Do not allow ECF for MP3 if buoyant force omitted.

Examiners report

Only those candidates who forgot to include the buoyant force missed marks here.

c(i).

[Maximum mark: 2]

20N.2.SL.TZ0.12c(i)

(c(i))

Determine the force exerted by the spring on the sphere when the sphere is at rest.

[2]

Markscheme

$F = m g - ρ V g$

OR

$F = (0.0126 \times 9.81) - (915 \times 7.24 \times 10^{- 6} \times 9.81)$ ✓

$F = 5.86 \times 10^{- 2} « N »$ ✓

Accept use of $g = 10$ leading to $F = 6.0 \times 10^{- 2} N$

Examiners report

Continuing from b, most candidates scored full marks.

a.

[Maximum mark: 2]

20N.2.SL.TZ0.14a

(a)

The astronomical unit ( $AU$ ) and light year ( $ly$ ) are convenient measures of distance in astrophysics. Define each unit.

$AU$ :

$ly$ :

[2]

Markscheme

$A U$ : «average» distance from the Earth to the Sun ✓

$l y$ : distance light travels in one year ✓

a.

[Maximum mark: 1]

20N.2.SL.TZ0.15a

(a)

Show that the apparent brightness $b \propto \frac{A T^{4}}{d^{2}}$ , where $d$ is the distance of the object from Earth, $T$ is the surface temperature of the object and $A$ is the surface area of the object.

[1]

Markscheme

substitution of $L = σ A T^{4}$ into $b = \frac{L}{4 {πd}^{2}}$ giving $b = \frac{σ A T^{4}}{4 {πd}^{2}}$

Removal of constants $σ$ and $4 π$ is optional

b.

[Maximum mark: 2]

20N.2.SL.TZ0.15b

(b)

Two of the brightest objects in the night sky seen from Earth are the planet Venus and the star Sirius. Explain why the equation $b \propto \frac{A T^{4}}{d^{2}}$ is applicable to Sirius but not to Venus.

[2]

Markscheme

equation applies to Sirius/stars that are luminous/emit light «from fusion» ✓

but Venus reflects the Sun’s light/does not emit light «from fusion» ✓

OWTTE

a(i).

[Maximum mark: 1]

20N.2.SL.TZ0.1a(i)

(a(i))

State the value of the resultant force on the aircraft when hovering.

[1]

Markscheme

zero ✓

a(ii).

[Maximum mark: 2]

20N.2.SL.TZ0.1a(ii)

(a(ii))

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

[2]

Markscheme

Blades exert a downward force on the air ✓

air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓

Downward direction required for MP1.

a(iii).

[Maximum mark: 3]

20N.2.SL.TZ0.1a(iii)

(a(iii))

Determine $v$ . State your answer to an appropriate number of significant figures.

[3]

Markscheme

«lift force/change of momentum in one second» $= 1.7 v$ ✓

$1.7 v = (0.95 + 0.45) \times 9.81$ ✓

$v = 8.1 « {ms}^{- 1} »$ AND answer expressed to $2$ sf only ✓

Allow $8.2$ from $g = 10 m s^{- 2}$ .

a(iv).

[Maximum mark: 2]

20N.2.SL.TZ0.1a(iv)

(a(iv))

Calculate the power transferred to the air by the aircraft.

[2]

Markscheme

ALTERNATIVE 1

power $« = rate of energy transfer to the air = \frac{1}{2} \frac{∆ m}{∆ t} v^{2} » = \frac{1}{2} \times 1.7 \times 8 . 1^{2}$ ✓

$= 56 « W »$ ✓

ALTERNATIVE 2

Power $« = Force \times v ave » = (0.95 + 0.45) \times 9.81 \times \frac{8.1}{2}$ ✓

$= 56 « W »$ ✓

Examiners report

HL only. It was common to see answers that neglected to average the velocity and consequently arrived at an answer twice the size of the correct one. This was awarded 1 of the 2 marks.

b.

[Maximum mark: 2]

20N.2.SL.TZ0.1b

(b)

The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.

[2]

Markscheme

vertical force = lift force – weight OR $= 0.45 \times 9.81$ OR $= 4.4 « N »$ ✓

acceleration $= \frac{0.45 \times 9.81}{0.95} = 4.6 « {ms}^{- 2} »$ ✓

3.

[Maximum mark: 8]

20N.2.SL.TZ0.3

A sample of vegetable oil, initially in the liquid state, is placed in a freezer that transfers thermal energy from the sample at a constant rate. The graph shows how temperature $T$ of the sample varies with time $t$ .

The following data are available.

Mass of the sample $= 0.32 kg$
Specific latent heat of fusion of the oil $= 130 kJ {kg}^{- 1}$
Rate of thermal energy transfer $= 15 W$

(a(i))

Calculate the thermal energy transferred from the sample during the first $30$ minutes.

[1]

Markscheme

$« 15 \times 30 \times 60 » = 27000 « J »$ ✓

(a(ii))

Estimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.

[2]

Markscheme

$27 \times 10^{3} = 0.32 \times c \times (290 - 250)$ OR $2100$ ✓

$J {kg}^{- 1} K^{- 1}$ OR $J {kg}^{- 1}^{0} C^{- 1}$ ✓

Allow any appropriate unit that is $\frac{e n e r g y}{m a s s \times t e r m p e r a t u r e}$

(b)

The sample begins to freeze during the thermal energy transfer. Explain, in terms of the molecular model of matter, why the temperature of the sample remains constant during freezing.

[3]

Markscheme

«intermolecular» bonds are formed during freezing ✓

bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓

«average random» KE of the molecules does not decrease/change ✓

temperature is related to «average» KE of the molecules «hence unchanged» ✓

To award MP3 or MP4 molecules/particles/atoms must be mentioned.

(c)

Calculate the mass of the oil that remains unfrozen after $60$ minutes.

[2]

Markscheme

mass of frozen oil $« = \frac{27 \times 10^{3}}{130 \times 10^{3}} » = 0.21 « kg »$ ✓

unfrozen mass $« = 0.32 - 0.21 » = 0.11 « kg »$ ✓

a(i).

[Maximum mark: 1]

20N.2.SL.TZ0.3a(i)

(a(i))

Calculate the thermal energy transferred from the sample during the first $30$ minutes.

[1]

Markscheme

$« 15 \times 30 \times 60 » = 27000 « J »$ ✓

a(ii).

[Maximum mark: 2]

20N.2.SL.TZ0.3a(ii)

(a(ii))

Estimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.

[2]

Markscheme

$27 \times 10^{3} = 0.32 \times c \times (290 - 250)$ OR $2100$ ✓

$J {kg}^{- 1} K^{- 1}$ OR $J {kg}^{- 1}^{0} C^{- 1}$ ✓

Allow any appropriate unit that is $\frac{e n e r g y}{m a s s \times t e r m p e r a t u r e}$

b.

[Maximum mark: 3]

20N.2.SL.TZ0.3b

(b)

The sample begins to freeze during the thermal energy transfer. Explain, in terms of the molecular model of matter, why the temperature of the sample remains constant during freezing.

[3]

Markscheme

«intermolecular» bonds are formed during freezing ✓

bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓

«average random» KE of the molecules does not decrease/change ✓

temperature is related to «average» KE of the molecules «hence unchanged» ✓

To award MP3 or MP4 molecules/particles/atoms must be mentioned.

c.

[Maximum mark: 2]

20N.2.SL.TZ0.3c

(c)

Calculate the mass of the oil that remains unfrozen after $60$ minutes.

[2]

Markscheme

mass of frozen oil $« = \frac{27 \times 10^{3}}{130 \times 10^{3}} » = 0.21 « kg »$ ✓

unfrozen mass $« = 0.32 - 0.21 » = 0.11 « kg »$ ✓

5.

[Maximum mark: 8]

20N.2.SL.TZ0.5

The graph shows how current $I$ varies with potential difference $V$ across a component X.

Component X and a cell of negligible internal resistance are placed in a circuit.

A variable resistor R is connected in series with component X. The ammeter reads $20 mA$ .

Component X and the cell are now placed in a potential divider circuit.

(a)

Outline why component X is considered non-ohmic.

[1]

Markscheme

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

(b(i))

Determine the resistance of the variable resistor.

[3]

Markscheme

ALTERNATIVE 1

voltage across X $= 2.3 « V »$ ✓

voltage across R $« = 4.0 - 2.3 » = 1.7 « V »$ ✓

resistance of variable resistor $« = \frac{1.7}{0.020} » = 85 « Ω »$ ✓

ALTERNATIVE 2

overall resistance $« = \frac{4.0}{0.020} » = 200 « Ω »$ ✓

resistance of X $« = \frac{2.3}{0.020} » = 115 « Ω »$ ✓

resistance of variable resistor $« = 200 - 115 » = 85 « Ω »$ ✓

(b(ii))

Calculate the power dissipated in the circuit.

[1]

Markscheme

power $« = 4.0 \times 0.020 » = 0.080 « W »$ ✓

(c(i))

State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.

[1]

Markscheme

from $0$ to $60 mA$ ✓

(c(ii))

Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).

[2]

Markscheme

allows zero current through component X / potential divider arrangement ✓

provides greater range «of current through component X» ✓

a.

[Maximum mark: 1]

20N.2.SL.TZ0.5a

(a)

Outline why component X is considered non-ohmic.

[1]

Markscheme

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

b(i).

[Maximum mark: 3]

20N.2.SL.TZ0.5b(i)

(b(i))

Determine the resistance of the variable resistor.

[3]

Markscheme

ALTERNATIVE 1

voltage across X $= 2.3 « V »$ ✓

voltage across R $« = 4.0 - 2.3 » = 1.7 « V »$ ✓

resistance of variable resistor $« = \frac{1.7}{0.020} » = 85 « Ω »$ ✓

ALTERNATIVE 2

overall resistance $« = \frac{4.0}{0.020} » = 200 « Ω »$ ✓

resistance of X $« = \frac{2.3}{0.020} » = 115 « Ω »$ ✓

resistance of variable resistor $« = 200 - 115 » = 85 « Ω »$ ✓

Examiners report

This question produced a mixture of answers from the 2 alternatives given in the markscheme. As a minimum, many candidates were able to score a mark for the overall resistance of the circuit.

b(ii).

[Maximum mark: 1]

20N.2.SL.TZ0.5b(ii)

(b(ii))

Calculate the power dissipated in the circuit.

[1]

Markscheme

power $« = 4.0 \times 0.020 » = 0.080 « W »$ ✓

Examiners report

A straightforward calculation question that most candidates answered correctly.

c(i).

[Maximum mark: 1]

20N.2.SL.TZ0.5c(i)

(c(i))

State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.

[1]

Markscheme

from $0$ to $60 mA$ ✓

Examiners report

Surprisingly a significant number of candidates had difficulty with this. Answers of 20 mA and 4 V were often seen.

c(ii).

[Maximum mark: 2]

20N.2.SL.TZ0.5c(ii)

(c(ii))

Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).

[2]

Markscheme

allows zero current through component X / potential divider arrangement ✓

provides greater range «of current through component X» ✓

a.

[Maximum mark: 1]

20N.2.SL.TZ0.7a

(a)

Outline why the cylinder performs simple harmonic motion when released.

[1]

Markscheme

the «restoring» force/acceleration is proportional to displacement ✓

Allow use of symbols i.e. $F \propto - x$ or $a \propto - x$

Examiners report

This was well answered with candidates gaining credit for answers in words or symbols.

b.

[Maximum mark: 2]

20N.2.SL.TZ0.7b

(b)

The mass of the cylinder is $118 kg$ and the cross-sectional area of the cylinder is $2.29 \times 10^{- 1} m^{2}$ . The density of water is $1.03 \times 10^{3} kg m^{- 3}$ . Show that the angular frequency of oscillation of the cylinder is about $4.4 rad s^{- 1}$ .

[2]

Markscheme

Evidence of equating $m ω^{2} x = ρ A g x$ «to obtain $\frac{ρ A g}{m} = ω^{2}$ » ✓

$ω = \sqrt{\frac{1.03 \times 10^{3} \times 2.29 \times 10^{- 1} \times 9.81}{118}}$ OR $4.43 « rad s^{- 1} »$ ✓

Answer to at least $3$ s.f.

Examiners report

Again, very well answered.

c(ii).

[Maximum mark: 2]

20N.2.SL.TZ0.7c(ii)

(c(ii))

Draw, on the axes, the graph to show how the kinetic energy of the cylinder varies with time during one period of oscillation $T$ .

[2]

Markscheme

energy never negative ✓

correct shape with two maxima ✓

Examiners report

Most candidates answered with a graph that was only positive so scored the first mark.

a.

[Maximum mark: 1]

20N.2.SL.TZ0.a

Outline why component X is considered non-ohmic.

[1]

Markscheme

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

Examiners report

Most answers that didn't score simply referred to the shape of the graph without any explanation as to what this meant to the relationship between the variables.

a(i).

[Maximum mark: 1]

20N.2.SL.TZ0.a(i)

State the value of the resultant force on the aircraft when hovering.

[1]

Markscheme

zero ✓

Examiners report

This was generally answered well with the most common incorrect answer being the weight of the aircraft and package. The question uses the command term 'state' which indicates that the answer requires no working.

a(ii).

[Maximum mark: 2]

20N.2.SL.TZ0.a(ii)

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

[2]

Markscheme

Blades exert a downward force on the air ✓

air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓

Downward direction required for MP1.

Examiners report

The question required candidates to apply Newton's third law to a specific situation. Candidates who had learned the 'action and reaction' version of Newton's third law generally did less well than those who had learned a version describing 'object A exerting a force on object B' etc. Some answers lacked detail of what was exerting the force and in which direction.

a(iii).

[Maximum mark: 3]

20N.2.SL.TZ0.a(iii)

Determine $v$ . State your answer to an appropriate number of significant figures.

[3]

Markscheme

«lift force/change of momentum in one second» $= 1.7 v$ ✓

$1.7 v = (0.95 + 0.45) \times 9.81$ ✓

$v = 8.1 « {ms}^{- 1} »$ AND answer expressed to $2$ sf only ✓

Allow $8.2$ from $g = 10 m s^{- 2}$ .

a(iv).

[Maximum mark: 2]

20N.2.SL.TZ0.a(iv)

Calculate the power transferred to the air by the aircraft.

[2]

Markscheme

ALTERNATIVE 1

power $« = rate of energy transfer to the air = \frac{1}{2} \frac{∆ m}{∆ t} v^{2} » = \frac{1}{2} \times 1.7 \times 8 . 1^{2}$ ✓

$= 56 « W »$ ✓

ALTERNATIVE 2

Power $« = Force \times v ave » = (0.95 + 0.45) \times 9.81 \times \frac{8.1}{2}$ ✓

$= 56 « W »$ ✓

Examiners report

HL only. It was common to see answers that neglected to average the velocity and consequently arrived at an answer twice the size of the correct one. This was awarded 1 of the 2 marks.

b.

[Maximum mark: 2]

20N.2.SL.TZ0.b

The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.

[2]

Markscheme

vertical force = lift force – weight OR $= 0.45 \times 9.81$ OR $= 4.4 « N »$ ✓

acceleration $= \frac{0.45 \times 9.81}{0.95} = 4.6 « {ms}^{- 2} »$ ✓

b(i).

[Maximum mark: 3]

20N.2.SL.TZ0.b(i)

Determine the resistance of the variable resistor.

[3]

Markscheme

ALTERNATIVE 1

voltage across X $= 2.3 « V »$ ✓

voltage across R $« = 4.0 - 2.3 » = 1.7 « V »$ ✓

resistance of variable resistor $« = \frac{1.7}{0.020} » = 85 « Ω »$ ✓

ALTERNATIVE 2

overall resistance $« = \frac{4.0}{0.020} » = 200 « Ω »$ ✓

resistance of X $« = \frac{2.3}{0.020} » = 115 « Ω »$ ✓

resistance of variable resistor $« = 200 - 115 » = 85 « Ω »$ ✓

b(ii).

[Maximum mark: 1]

20N.2.SL.TZ0.b(ii)

Calculate the power dissipated in the circuit.

[1]

Markscheme

power $« = 4.0 \times 0.020 » = 0.080 « W »$ ✓

c.

[Maximum mark: 2]

20N.2.SL.TZ0.c

Calculate the mass of the oil that remains unfrozen after $60$ minutes.

[2]

Markscheme

mass of frozen oil $« = \frac{27 \times 10^{3}}{130 \times 10^{3}} » = 0.21 « kg »$ ✓

unfrozen mass $« = 0.32 - 0.21 » = 0.11 « kg »$ ✓

c(i).

[Maximum mark: 2]

20N.2.SL.TZ0.c(i)

Determine the force exerted by the spring on the sphere when the sphere is at rest.

[2]

Markscheme

$F = m g - ρ V g$

OR

$F = (0.0126 \times 9.81) - (915 \times 7.24 \times 10^{- 6} \times 9.81)$ ✓

$F = 5.86 \times 10^{- 2} « N »$ ✓

Accept use of $g = 10$ leading to $F = 6.0 \times 10^{- 2} N$

Examiners report

Continuing from b, most candidates scored full marks.

c(ii).

[Maximum mark: 3]

20N.2.SL.TZ0.c(ii)

Slider S of the potential divider is positioned so that the ammeter reads $20 mA$ . Explain, without further calculation, any difference in the power transferred by the potential divider arrangement over the arrangement in (b).

[3]

Markscheme

ALTERNATIVE 1

current from the cell is greater «than $20 mA$ » ✓

because some of the current must flow through section SQ of the potentiometer ✓

overall power greater «than in part (b)» ✓

ALTERNATIVE 2

total/overall resistance decreases ✓

because SQ and X are in parallel ✓

overall power greater «than in part (b)» ✓

Allow the reverse argument.

Examiners report

HL only. This question challenged candidate's ability to describe clearly the changes in an electrical circuit. It revealed many misconceptions about the nature of electrical current and potential difference, of those who did have a grasp of what was going on the explanations often missed the second point in each of the markscheme alternatives as detail was missed about where the current was flowing or what was in parallel with what.

3.

[Maximum mark: 1]

21M.1A.SL.TZ1.3

A large stone is dropped from a tall building. What is correct about the speed of the stone after 1 s?

A. It is decreasing at increasing rate.

B. It is decreasing at decreasing rate.

C. It is increasing at increasing rate.

D. It is increasing at decreasing rate.

[1]

Markscheme

D

30.

[Maximum mark: 1]

21M.1A.SL.TZ1.30

In a simple climate model for a planet, the incoming intensity is 400 W m⁻² and the radiated intensity is 300 W m⁻².

The temperature of the planet is constant. What are the reflected intensity from the planet and the albedo of the planet?

[1]

Markscheme

A

4.

[Maximum mark: 1]

21M.1A.SL.TZ1.4

The graph shows how the position of an object varies with time in the interval from 0 to 3 s.

At which point does the instantaneous speed of the object equal its average speed over the interval from 0 to 3 s?

[1]

Markscheme

C

5.

[Maximum mark: 1]

21M.1A.SL.TZ1.5

A car takes 20 minutes to climb a hill at constant speed. The mass of the car is 1200 kg and the car gains gravitational potential energy at a rate of 6.0 kW. Take the acceleration of gravity to be 10 m s⁻². What is the height of the hill?

A. 0.6 m

B. 10 m

C. 600 m

D. 6000 m

[1]

Markscheme

C

3.

[Maximum mark: 1]

21M.1A.SL.TZ2.3

The minute hand of a clock hanging on a vertical wall has length $L = 30 cm.$

The minute hand is observed pointing at 12 and then again 30 minutes later when the minute hand is pointing at 6.

What is the average velocity and average speed of point P on the minute hand during this time interval?

[1]

Markscheme

A

30.

[Maximum mark: 1]

21M.1A.SL.TZ2.30

What is the main role of carbon dioxide in the greenhouse effect?

A. It absorbs incoming radiation from the Sun.

B. It absorbs outgoing radiation from the Earth.

C. It reflects incoming radiation from the Sun.

D. It reflects outgoing radiation from the Earth.

[1]

Markscheme

B

4.

[Maximum mark: 1]

21M.1A.SL.TZ2.4

A projectile is launched at an angle above the horizontal with a horizontal component of velocity $V_{h}$ and a vertical component of velocity $V_{v}$ . Air resistance is negligible. Which graphs show the variation with time of $V_{h}$ and of $V_{v}$ ?

[1]

Markscheme

D

40.

[Maximum mark: 1]

21M.1A.SL.TZ2.40

An electron of non-relativistic speed $v$ interacts with an atom. All the energy of the electron is transferred to an emitted photon of frequency $f$ . An electron of speed $2 v$ now interacts with the same atom and all its energy is transmitted to a second photon. What is the frequency of the second photon?

A. $\frac{f}{4}$

B. $\frac{f}{2}$

C. $2 f$

D. $4 f$

[1]

Markscheme

D

1.

[Maximum mark: 11]

21M.2.SL.TZ1.1

Two players are playing table tennis. Player A hits the ball at a height of 0.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. The initial speed of the ball is 12.0 m s⁻¹ horizontally. Assume that air resistance is negligible.

The ball bounces and then reaches a peak height of 0.18 m above the table with a horizontal speed of 10.5 m s⁻¹. The mass of the ball is 2.7 g.

(a)

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

[1]

Markscheme

t = « $\sqrt{\frac{2 d}{g}}$ =» 0.22 «s»
OR

t = $\sqrt{\frac{2 \times 0.24}{9.8}}$ ✓

Answer to 2 or more significant figures or formula with variables replaced by correct values.

(b)

Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity v_v of the ball until it reaches the table surface. Take g to be +10 m s⁻².

[2]

Markscheme

increasing straight line from zero up to 0.2 s in x-axis ✓

with gradient = 10 ✓

(c)

The net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.

Show that the ball will go over the net.

[3]

Markscheme

ALTERNATIVE 1

$t = \frac{1.37}{12} =$ «0.114 s» ✓

$y = \frac{1}{2} \times 10 \times 0 . 114^{2} = 0.065$ m ✓

so (0.24 − 0.065) = 0.175 > 0.15 OR 0.065 < (0.24 − 0.15) «so it goes over the net» ✓

ALTERNATIVE 2

«0.24 − 0.15 = 0.09 = $\frac{1}{2} \times 10 \times t^{2}$ so» t = 0.134 s ✓

0.134 × 12 = 1.6 m ✓

1.6 > 1.37 «so ball passed the net already» ✓

Allow use of g = 9.8.

(d.i)

Determine the kinetic energy of the ball immediately after the bounce.

[2]

Markscheme

ALTERNATIVE 1

KE = $\frac{1}{2}$ mv² + mgh = $\frac{1}{2}$ 0.0027 ×10.5² + 0.0027 × 9.8 × 0.18 ✓

0.15 «J» ✓

ALTERNATIVE 2

Use of v_x = 10.5 AND v_y= 1.88 to get v = « $\sqrt{10 . 5^{2} + 1 . 88^{2}}$ » = 10.67 «m s⁻¹» ✓

KE = $\frac{1}{2}$ × 0.0027 × 10.67² = 0.15 «J» ✓

(d.ii)

Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

[3]

Markscheme

$Δ v = 21$ «m s⁻¹» ✓

$F = \frac{0.0027 \times 21}{0.01}$

OR

5.67 «N» ✓

any answer to 2 significant figures «N» ✓

a.

[Maximum mark: 1]

21M.2.SL.TZ1.1a

(a)

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

[1]

Markscheme

t = « $\sqrt{\frac{2 d}{g}}$ =» 0.22 «s»
OR

t = $\sqrt{\frac{2 \times 0.24}{9.8}}$ ✓

Answer to 2 or more significant figures or formula with variables replaced by correct values.

b.

[Maximum mark: 2]

21M.2.SL.TZ1.1b

(b)

Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity v_v of the ball until it reaches the table surface. Take g to be +10 m s⁻².

[2]

Markscheme

increasing straight line from zero up to 0.2 s in x-axis ✓

with gradient = 10 ✓

c.

[Maximum mark: 3]

21M.2.SL.TZ1.1c

(c)

The net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.

Show that the ball will go over the net.

[3]

Markscheme

ALTERNATIVE 1

$t = \frac{1.37}{12} =$ «0.114 s» ✓

$y = \frac{1}{2} \times 10 \times 0 . 114^{2} = 0.065$ m ✓

so (0.24 − 0.065) = 0.175 > 0.15 OR 0.065 < (0.24 − 0.15) «so it goes over the net» ✓

ALTERNATIVE 2

«0.24 − 0.15 = 0.09 = $\frac{1}{2} \times 10 \times t^{2}$ so» t = 0.134 s ✓

0.134 × 12 = 1.6 m ✓

1.6 > 1.37 «so ball passed the net already» ✓

Allow use of g = 9.8.

d.i.

[Maximum mark: 2]

21M.2.SL.TZ1.1d.i

(d.i)

Determine the kinetic energy of the ball immediately after the bounce.

[2]

Markscheme

ALTERNATIVE 1

KE = $\frac{1}{2}$ mv² + mgh = $\frac{1}{2}$ 0.0027 ×10.5² + 0.0027 × 9.8 × 0.18 ✓

0.15 «J» ✓

ALTERNATIVE 2

Use of v_x = 10.5 AND v_y= 1.88 to get v = « $\sqrt{10 . 5^{2} + 1 . 88^{2}}$ » = 10.67 «m s⁻¹» ✓

KE = $\frac{1}{2}$ × 0.0027 × 10.67² = 0.15 «J» ✓

d.ii.

[Maximum mark: 3]

21M.2.SL.TZ1.1d.ii

(d.ii)

Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

[3]

Markscheme

$Δ v = 21$ «m s⁻¹» ✓

$F = \frac{0.0027 \times 21}{0.01}$

OR

5.67 «N» ✓

any answer to 2 significant figures «N» ✓

a.

[Maximum mark: 2]

21M.2.SL.TZ1.2a

(a)

Explain why a centripetal force is needed for the planet to be in a circular orbit.

[2]

Markscheme

«circular motion» involves a changing velocity ✓

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

there must be a force/acceleration towards centre/star ✓

without a centripetal force the planet will move in a straight line ✓

b.

[Maximum mark: 1]

21M.2.SL.TZ1.2b

(b)

Calculate the value of the centripetal force.

[1]

Markscheme

$F = \frac{(6.67 \times 10^{- 11}) (8 \times 10^{24}) (3.2 \times 10^{30})}{(4.4 \times 10^{10})^{2}} = 8.8 \times 10^{23}$ «N» ✓

3.

[Maximum mark: 12]

21M.2.SL.TZ1.3

A mass of 1.0 kg of water is brought to its boiling point of 100 °C using an electric heater of power 1.6 kW.

A mass of 0.86 kg of water remains after it has boiled for 200 s.

The electric heater has two identical resistors connected in parallel.

The circuit transfers 1.6 kW when switch A only is closed. The external voltage is 220 V.

(a.i)

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

[2]

Markscheme

E_k = « $\frac{3}{2} (1.38 \times 10^{- 23}) (373)$ » = $7.7 \times 10^{- 21}$ «J» ✓

v = « $\sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 6.02 \times 10^{23} \times 373}{0.018}}$ » = 720 «m s⁻¹» ✓

(a.ii)

State one assumption of the kinetic model of an ideal gas.

[1]

Markscheme

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles ✓

Accept reference to atoms/molecules for “particle”

(b.i)

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

[2]

Markscheme

«mL = P t» so « $L = \frac{1600 \times 200}{0.14}$ » = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹✓

(b.ii)

Explain why the temperature of water remains at 100 °C during this time.

[1]

Markscheme

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

Accept reference to atoms/molecules for “particle”

(c)

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg⁻¹ K⁻¹
Specific heat capacity of water = 4.2 kJ kg⁻¹ K⁻¹

[3]

Markscheme

use of mcΔT ✓

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

T_eq = 85.69«°C» ≅ 86«°C» ✓

Accept T_eq in Kelvin (359 K).

(d.i)

Show that each resistor has a resistance of about 30 Ω.

[1]

Markscheme

$P = \frac{v^{2}}{R} so \frac{220^{2}}{1600} so R = 30.25$ «Ω» ✓

Must see either the substituted values OR a value for R to at least three s.f.

(d.ii)

Calculate the power transferred by the heater when both switches are closed.

[2]

Markscheme

use of parallel resistors addition so R_eq = 15 «Ω» ✓

P = 3200 «W» ✓

a.i.

[Maximum mark: 2]

21M.2.SL.TZ1.3a.i

(a.i)

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

[2]

Markscheme

E_k = « $\frac{3}{2} (1.38 \times 10^{- 23}) (373)$ » = $7.7 \times 10^{- 21}$ «J» ✓

v = « $\sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 6.02 \times 10^{23} \times 373}{0.018}}$ » = 720 «m s⁻¹» ✓

a.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.3a.ii

(a.ii)

State one assumption of the kinetic model of an ideal gas.

[1]

Markscheme

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles ✓

Accept reference to atoms/molecules for “particle”

b.i.

[Maximum mark: 2]

21M.2.SL.TZ1.3b.i

(b.i)

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

[2]

Markscheme

«mL = P t» so « $L = \frac{1600 \times 200}{0.14}$ » = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹✓

b.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.3b.ii

(b.ii)

Explain why the temperature of water remains at 100 °C during this time.

[1]

Markscheme

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

Accept reference to atoms/molecules for “particle”

c.

[Maximum mark: 3]

21M.2.SL.TZ1.3c

(c)

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg⁻¹ K⁻¹
Specific heat capacity of water = 4.2 kJ kg⁻¹ K⁻¹

[3]

Markscheme

use of mcΔT ✓

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

T_eq = 85.69«°C» ≅ 86«°C» ✓

Accept T_eq in Kelvin (359 K).

d.i.

[Maximum mark: 1]

21M.2.SL.TZ1.3d.i

(d.i)

Show that each resistor has a resistance of about 30 Ω.

[1]

Markscheme

$P = \frac{v^{2}}{R} so \frac{220^{2}}{1600} so R = 30.25$ «Ω» ✓

Must see either the substituted values OR a value for R to at least three s.f.

d.ii.

[Maximum mark: 2]

21M.2.SL.TZ1.3d.ii

(d.ii)

Calculate the power transferred by the heater when both switches are closed.

[2]

Markscheme

use of parallel resistors addition so R_eq = 15 «Ω» ✓

P = 3200 «W» ✓

6.

[Maximum mark: 5]

21M.2.SL.TZ1.6

On a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger.

The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm.

(a)

Outline how a standing wave is produced on the string.

[2]

Markscheme

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

superposition/interference of incident and reflected waves ✓

the superposition of the reflections is reinforced only for certain wavelengths ✓

(b.i)

Show that the speed of the wave on the string is about 240 m s⁻¹.

[2]

Markscheme

$λ = 2 l = 2 \times 0.62 = « 1.24 m »$ ✓

$v = f λ = 195 \times 1.24 = 242 « {m s}^{- 1} »$ ✓

Answer must be to 3 or more sf or working shown for MP2.

(b.ii)

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

[1]

Markscheme

straight line through origin with negative gradient ✓

a.

[Maximum mark: 2]

21M.2.SL.TZ1.6a

(a)

Outline how a standing wave is produced on the string.

[2]

Markscheme

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

superposition/interference of incident and reflected waves ✓

the superposition of the reflections is reinforced only for certain wavelengths ✓

b.i.

[Maximum mark: 2]

21M.2.SL.TZ1.6b.i

(b.i)

Show that the speed of the wave on the string is about 240 m s⁻¹.

[2]

Markscheme

$λ = 2 l = 2 \times 0.62 = « 1.24 m »$ ✓

$v = f λ = 195 \times 1.24 = 242 « {m s}^{- 1} »$ ✓

Answer must be to 3 or more sf or working shown for MP2.

b.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.6b.ii

(b.ii)

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

[1]

Markscheme

straight line through origin with negative gradient ✓

7.

[Maximum mark: 4]

21M.2.SL.TZ1.7

Conservation of energy and conservation of momentum are two examples of conservation laws.

(a)

Outline the significance of conservation laws for physics.

[1]

Markscheme

they express fundamental principles of nature ✓

allow to model situations ✓

allow to calculate unknown variables ✓

allow to predict possible outcomes ✓

allow to predict missing quantities/particles ✓

allow comparison of different system states ✓

(b)

When a pi meson π- (du̅) and a proton (uud) collide, a possible outcome is a sigma baryon Σ⁰ (uds) and a kaon meson Κ⁰ (ds̅).

Apply three conservation laws to show that this interaction is possible.

[3]

Markscheme

three correct conservation laws listed ✓

at least one conservation law correctly demonstrated ✓

all three conservation laws correctly demonstrated ✓

a.

[Maximum mark: 1]

21M.2.SL.TZ1.7a

(a)

Outline the significance of conservation laws for physics.

[1]

Markscheme

they express fundamental principles of nature ✓

allow to model situations ✓

allow to calculate unknown variables ✓

allow to predict possible outcomes ✓

allow to predict missing quantities/particles ✓

allow comparison of different system states ✓

a.

[Maximum mark: 2]

21M.2.SL.TZ1.8a

(a)

Outline how a standing wave is produced on the string.

[2]

Markscheme

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

superposition/interference of incident and reflected waves ✓

the superposition of the reflections is reinforced only for certain wavelengths ✓

b.i.

[Maximum mark: 2]

21M.2.SL.TZ1.8b.i

(b.i)

Show that the speed of the wave on the string is about 240 m s⁻¹.

[2]

Markscheme

$λ = 2 l = 2 \times 0.62 = « 1.24 m »$ ✓

$v = f λ = 195 \times 1.24 = 242 « {m s}^{- 1} »$ ✓

Answer must be to 3 or more sf or working shown for MP2.

b.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.8b.ii

(b.ii)

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

[1]

Markscheme

straight line through origin with negative gradient ✓

c.

[Maximum mark: 1]

21M.2.SL.TZ1.8c

(c)

The string is made to vibrate in its third harmonic. State the distance between consecutive nodes.

[1]

Markscheme

$\frac{62}{3} = 21 « cm »$ ✓

a.

[Maximum mark: 1]

21M.2.SL.TZ1.a

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

[1]

Markscheme

t = « $\sqrt{\frac{2 d}{g}}$ =» 0.22 «s»
OR

t = $\sqrt{\frac{2 \times 0.24}{9.8}}$ ✓

Answer to 2 or more significant figures or formula with variables replaced by correct values.

a.i.

[Maximum mark: 2]

21M.2.SL.TZ1.a.i

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

[2]

Markscheme

E_k = « $\frac{3}{2} (1.38 \times 10^{- 23}) (373)$ » = $7.7 \times 10^{- 21}$ «J» ✓

v = « $\sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 6.02 \times 10^{23} \times 373}{0.018}}$ » = 720 «m s⁻¹» ✓

a.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.a.ii

State one assumption of the kinetic model of an ideal gas.

[1]

Markscheme

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles ✓

Accept reference to atoms/molecules for “particle”

b.

[Maximum mark: 2]

21M.2.SL.TZ1.b

Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity v_v of the ball until it reaches the table surface. Take g to be +10 m s⁻².

[2]

Markscheme

increasing straight line from zero up to 0.2 s in x-axis ✓

with gradient = 10 ✓

b.i.

[Maximum mark: 2]

21M.2.SL.TZ1.b.i

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

[2]

Markscheme

«mL = P t» so « $L = \frac{1600 \times 200}{0.14}$ » = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹✓

b.ii.

[Maximum mark: 1]

21M.2.SL.TZ1.b.ii

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

[1]

Markscheme

straight line through origin with negative gradient ✓

c.

[Maximum mark: 3]

21M.2.SL.TZ1.c

The net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.

Show that the ball will go over the net.

[3]

Markscheme

ALTERNATIVE 1

$t = \frac{1.37}{12} =$ «0.114 s» ✓

$y = \frac{1}{2} \times 10 \times 0 . 114^{2} = 0.065$ m ✓

so (0.24 − 0.065) = 0.175 > 0.15 OR 0.065 < (0.24 − 0.15) «so it goes over the net» ✓

ALTERNATIVE 2

«0.24 − 0.15 = 0.09 = $\frac{1}{2} \times 10 \times t^{2}$ so» t = 0.134 s ✓

0.134 × 12 = 1.6 m ✓

1.6 > 1.37 «so ball passed the net already» ✓

Allow use of g = 9.8.

d.i.

[Maximum mark: 1]

21M.2.SL.TZ1.d.i

Show that each resistor has a resistance of about 30 Ω.

[1]

Markscheme

$P = \frac{v^{2}}{R} so \frac{220^{2}}{1600} so R = 30.25$ «Ω» ✓

Must see either the substituted values OR a value for R to at least three s.f.

d.ii.

[Maximum mark: 3]

21M.2.SL.TZ1.d.ii

Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

[3]

Markscheme

$Δ v = 21$ «m s⁻¹» ✓

$F = \frac{0.0027 \times 21}{0.01}$

OR

5.67 «N» ✓

any answer to 2 significant figures «N» ✓

b.i.

[Maximum mark: 2]

21M.2.SL.TZ2.1b.i

(b.i)

The ball leaves the ground at an angle of 22°. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m. Calculate the time taken for the ball to reach the wall.

[2]

Markscheme

$horizontal speed = 19 \times \cos 22 « = 17.6 {m s}^{- 1} »$ ✓

$time = « \frac{distance}{speed} = \frac{11}{19 \cos 22} = » 0.62 « s »$ ✓

Allow ECF for MP2

b.ii.

[Maximum mark: 3]

21M.2.SL.TZ2.1b.ii

(b.ii)

The top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.

[3]

Markscheme

$initial vertical speed = 19 \times \sin 22 « = 7.1 {m s}^{- 1} »$ ✓

$« 7.12 \times 0.624 - 0.5 \times 9.81 \times 0 . 624^{2} = » 2.5 « m »$ ✓

ball does not hit wall OR 2.5 «m» > 2.4 «m» ✓

Allow ECF from (b)(i) and from MP1

Allow g = 10 m s⁻²

b.

[Maximum mark: 3]

21M.2.SL.TZ2.b

Calculate the internal resistance of the photovoltaic cell for the maximum intensity condition using the model for the cell.

[3]

Markscheme

ALTERNATIVE 1
pd dropped across cell $= 6.5 « V »$ ✓

internal resistance $= \frac{6.5}{0.9}$ ✓

$7.2 « Ω »$ ✓

ALTERNATIVE 2

$ε = I (R + r)$ so $ε = V + I r$ ✓

$21.0 = 14.5 + 0.9 \times r$ ✓

$7.2 « Ω »$ ✓

Alternative solutions are possible

Award [3] marks for a bald correct answer

b.i.

[Maximum mark: 2]

21M.2.SL.TZ2.b.i

The ball leaves the ground at an angle of 22°. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m. Calculate the time taken for the ball to reach the wall.

[2]

Markscheme

$horizontal speed = 19 \times \cos 22 « = 17.6 {m s}^{- 1} »$ ✓

$time = « \frac{distance}{speed} = \frac{11}{19 \cos 22} = » 0.62 « s »$ ✓

Allow ECF for MP2

b.ii.

[Maximum mark: 3]

21M.2.SL.TZ2.b.ii

The top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.

[3]

Markscheme

$initial vertical speed = 19 \times \sin 22 « = 7.1 {m s}^{- 1} »$ ✓

$« 7.12 \times 0.624 - 0.5 \times 9.81 \times 0 . 624^{2} = » 2.5 « m »$ ✓

ball does not hit wall OR 2.5 «m» > 2.4 «m» ✓

Allow ECF from (b)(i) and from MP1

Allow g = 10 m s⁻²

b.iii.

[Maximum mark: 2]

21M.2.SL.TZ2.b.iii

Between the first and second positions of maximum loudness, the tube is raised through 0.37 m. The speed of sound in the air in the tube is 320 m s⁻¹. Determine the frequency of the sound emitted by the loudspeaker.

[2]

Markscheme

$λ = 0.74 « m »$ ✓

$f = « \frac{c}{λ} = \frac{320}{0.74} = » 430 « Hz »$ ✓

Allow ECF from MP1

3.

[Maximum mark: 1]

21N.1A.SL.TZ0.3

The graph shows the variation with time t of the velocity of an object.

What is the variation with time t of the acceleration of the object?

[1]

Markscheme

A

30.

[Maximum mark: 1]

21N.1A.SL.TZ0.30

Which is correct for a black-body radiator?

A. The power it emits from a unit surface area depends on the temperature only.

B. It has an albedo of 1.

C. It emits monochromatic radiation whose wavelength depends on the temperature only.

D. It emits radiation of equal intensity at all wavelengths.

[1]

Markscheme

A

4.

[Maximum mark: 1]

21N.1A.SL.TZ0.4

A ball is thrown vertically downwards with an initial speed of 4.0 m s⁻¹. The ball hits the ground with a speed of 16 m s⁻¹. Air resistance is negligible. What is the time of fall and what is the distance travelled by the ball?

[1]

Markscheme

D

5.

[Maximum mark: 1]

22M.1A.SL.TZ1.5

A stone is kicked horizontally at a speed of 1.5 m s⁻¹ from the edge of a cliff on one of Jupiter’s moons. It hits the ground 2.0 s later. The height of the cliff is 4.0 m. Air resistance is negligible.

What is the magnitude of the displacement of the stone?

A. 7.0 m

B. 5.0 m

C. 4.0 m

D. 3.0 m

[1]

Markscheme

B

Examiners report

This question was generally well answered by both HL and SL candidates and had a mid-range difficulty index (indicating an easier question). Option D was an effective distractor for candidates calculating the horizontal range rather than the displacement. Candidates are encouraged to read the questions carefully to ensure it is clear what each question is asking for.

3.

[Maximum mark: 1]

22M.1A.SL.TZ2.3

The road from city X to city Y is 1000 km long. The displacement is 800 km from X to Y.

What is the distance travelled from Y to X and the displacement from Y to X?

[1]

Markscheme

D

30.

[Maximum mark: 1]

22M.1A.SL.TZ2.30

The Sankey diagrams for a filament lamp and for an LED bulb are shown below.

What is the efficiency of the filament lamp and the LED bulb?

[1]

Markscheme

A

31.

[Maximum mark: 1]

22M.1A.SL.TZ2.31

A train is sounding its whistle when approaching a train station. Three statements about the sound received by a stationary observer at the station are:

I. The frequency received is higher than the frequency emitted by the train.
II. The wavelength received is longer than the wavelength emitted by the train.
III. The speed of the sound received is not affected by the motion of the train.

Which combination of statements is correct?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

[1]

Markscheme

B

4.

[Maximum mark: 1]

22M.1A.SL.TZ2.4

A car accelerates uniformly from rest to a velocity $v$ during time $t_{1}$ . It then continues at constant velocity $v$ from $t_{1}$ to time $t_{2}$ .

What is the total distance covered by the car in $t_{2}$ ?

A. $v t_{2}$

B.   $\frac{1}{2} v (t_{2} - t_{1}) + v t_{1}$

C.   $\frac{1}{2} v (t_{2} + t_{1})$

D.   $\frac{1}{2} v t_{1} + v (t_{2} - t_{1})$

[1]

Markscheme

D

5.

[Maximum mark: 1]

22M.1A.SL.TZ2.5

An object is sliding from rest down a frictionless inclined plane. The object slides 1.0 m during the first second.

What distance will the object slide during the next second?

A. 1.0 m

B. 2.0 m

C. 3.0 m

D. 4.9 m

[1]

Markscheme

C

Examiners report

The correct response, option C was the most popular chosen at HL but at SL significantly more candidates chose options A or B. The difficulty index of 21 and discrimination index of 0.27 at SL indicates that students found the question to be hard with lower discrimination between stronger and weaker candidates. It is felt that those who chose option A did not realise the block was accelerating down the slope, whereas those choosing B did but were unable to calculate the acceleration correctly.

1.

[Maximum mark: 15]

22M.2.SL.TZ1.1

A student uses a load to pull a box up a ramp inclined at 30°. A string of constant length and negligible mass connects the box to the load that falls vertically. The string passes over a pulley that runs on a frictionless axle. Friction acts between the base of the box and the ramp. Air resistance is negligible.

The load has a mass of 3.5 kg and is initially 0.95 m above the floor. The mass of the box is 1.5 kg.

The load is released and accelerates downwards.

(a)

Outline two differences between the momentum of the box and the momentum of the load at the same instant.

[2]

Markscheme

direction of motion is different / OWTTE ✓

mv / magnitude of momentum is different «even though v the same» ✓

Examiners report

Many students recognized the vector nature of momentum implied in the question, although some focused on the forces acting on each object rather than discussing the momentum.

(b)

The vertical acceleration of the load downwards is 2.4 m s⁻².

Calculate the tension in the string.

[2]

Markscheme

use of ma = mg − T «3.5 x 2.4 = 3.5g − T »

OR

T = 3.5(g − 2.4) ✓

26 «N» ✓

Accept 27 N from g = 10 m s⁻²

Examiners report

Some students simply calculated the net force acting on the load and did not recognize that this was not the tension force. Many set up a net force equation but had the direction of the forces backwards. This generally resulted from sloppy problem solving.

(c.i)

Show that the speed of the load when it hits the floor is about 2.1 m s⁻¹.

[2]

Markscheme

proper use of kinematic equation ✓

$\sqrt{(2 \times 2.4 \times 0.95)} = 2.14$ «m s⁻¹» ✓

Must see either the substituted values OR a value for v to at least three s.f. for MP2.

Examiners report

This was a "show that" questions, so examiners were looking for a clear equation leading to a clear substitution of values leading to an answer that had more significant digits than the given answer. Most candidates successfully selected the correct equation and showed a proper substitution. Some candidates started with an energy approach that needed modification as it clearly led to an incorrect solution. These responses did not receive full marks.

(c.ii)

The radius of the pulley is 2.5 cm. Calculate the angular speed of rotation of the pulley as the load hits the floor. State your answer to an appropriate number of significant figures.

[2]

Markscheme

use of $ω = \frac{v}{r}$ to give 84 «rad s⁻¹»

OR

$ω = 2.1 / 0.025$ to give 84 «rad s⁻¹» ✓

quoted to 2sf only✓

Examiners report

This SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.

(d)

After the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.

[4]

Markscheme

ALTERNATIVE 1

« $v^{2} = u^{2} + 2 a s \Rightarrow 0 = 2 . 1^{2} - 2 a \times 0.35$ » leading to a = 6.3 «m s^-2»

OR

« $x = 1 / 2 (u + v) t$ » leading to t = 0.33 « s » ✓

F_net = « $m a = 1.5 \times 6.3$ = » 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

ALTERNATIVE 2

kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)² = F_NET x 0.35 ✓

F_net = 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

Accept 1.95 N from g = 10 m s^-2.
Accept 2.42 N from u = 2.14 m s^-1.

Examiners report

Candidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.

(e)

The student then makes the ramp horizontal and applies a constant horizontal force to the box. The force is just large enough to start the box moving. The force continues to be applied after the box begins to move.

Explain, with reference to the frictional force acting, why the box accelerates once it has started to move.

[3]

Markscheme

static coefficient of friction > dynamic/kinetic coefficient of friction / μ_s > μ_k ✓

«therefore» force of dynamic/kinetic friction will be less than the force of static friction ✓

there will be a net / unbalanced forward force once in motion «which results in acceleration»

OR

reference to net F = ma ✓

Examiners report

This was an "explain" question, so examiners were looking for a clear line of discussion starting with a comparison of the coefficients of friction, leading to a comparison of the relative magnitudes of the forces of friction and ultimately the rise of a net force leading to an acceleration. Many candidates recognized that this was a question about the comparison between static and kinetic/dynamic friction but did not clearly specify which they were referring to in their responses. Some candidates clearly did not read the stem carefully as they referred to the mass being on an incline.

a.

[Maximum mark: 2]

22M.2.SL.TZ1.1a

(a)

Outline two differences between the momentum of the box and the momentum of the load at the same instant.

[2]

Markscheme

direction of motion is different / OWTTE ✓

mv / magnitude of momentum is different «even though v the same» ✓

Examiners report

Many students recognized the vector nature of momentum implied in the question, although some focused on the forces acting on each object rather than discussing the momentum.

b.

[Maximum mark: 2]

22M.2.SL.TZ1.1b

(b)

The vertical acceleration of the load downwards is 2.4 m s⁻².

Calculate the tension in the string.

[2]

Markscheme

use of ma = mg − T «3.5 x 2.4 = 3.5g − T »

OR

T = 3.5(g − 2.4) ✓

26 «N» ✓

Accept 27 N from g = 10 m s⁻²

Examiners report

Some students simply calculated the net force acting on the load and did not recognize that this was not the tension force. Many set up a net force equation but had the direction of the forces backwards. This generally resulted from sloppy problem solving.

c.i.

[Maximum mark: 2]

22M.2.SL.TZ1.1c.i

(c.i)

Show that the speed of the load when it hits the floor is about 2.1 m s⁻¹.

[2]

Markscheme

proper use of kinematic equation ✓

$\sqrt{(2 \times 2.4 \times 0.95)} = 2.14$ «m s⁻¹» ✓

Must see either the substituted values OR a value for v to at least three s.f. for MP2.

Examiners report

This was a "show that" questions, so examiners were looking for a clear equation leading to a clear substitution of values leading to an answer that had more significant digits than the given answer. Most candidates successfully selected the correct equation and showed a proper substitution. Some candidates started with an energy approach that needed modification as it clearly led to an incorrect solution. These responses did not receive full marks.

c.ii.

[Maximum mark: 2]

22M.2.SL.TZ1.1c.ii

(c.ii)

The radius of the pulley is 2.5 cm. Calculate the angular speed of rotation of the pulley as the load hits the floor. State your answer to an appropriate number of significant figures.

[2]

Markscheme

use of $ω = \frac{v}{r}$ to give 84 «rad s⁻¹»

OR

$ω = 2.1 / 0.025$ to give 84 «rad s⁻¹» ✓

quoted to 2sf only✓

Examiners report

This SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.

d.

[Maximum mark: 4]

22M.2.SL.TZ1.1d

(d)

After the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.

[4]

Markscheme

ALTERNATIVE 1

« $v^{2} = u^{2} + 2 a s \Rightarrow 0 = 2 . 1^{2} - 2 a \times 0.35$ » leading to a = 6.3 «m s^-2»

OR

« $x = 1 / 2 (u + v) t$ » leading to t = 0.33 « s » ✓

F_net = « $m a = 1.5 \times 6.3$ = » 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

ALTERNATIVE 2

kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)² = F_NET x 0.35 ✓

F_net = 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

Accept 1.95 N from g = 10 m s^-2.
Accept 2.42 N from u = 2.14 m s^-1.

Examiners report

Candidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.

e.

[Maximum mark: 3]

22M.2.SL.TZ1.1e

(e)

The student then makes the ramp horizontal and applies a constant horizontal force to the box. The force is just large enough to start the box moving. The force continues to be applied after the box begins to move.

Explain, with reference to the frictional force acting, why the box accelerates once it has started to move.

[3]

Markscheme

static coefficient of friction > dynamic/kinetic coefficient of friction / μ_s > μ_k ✓

«therefore» force of dynamic/kinetic friction will be less than the force of static friction ✓

there will be a net / unbalanced forward force once in motion «which results in acceleration»

OR

reference to net F = ma ✓

Examiners report

This was an "explain" question, so examiners were looking for a clear line of discussion starting with a comparison of the coefficients of friction, leading to a comparison of the relative magnitudes of the forces of friction and ultimately the rise of a net force leading to an acceleration. Many candidates recognized that this was a question about the comparison between static and kinetic/dynamic friction but did not clearly specify which they were referring to in their responses. Some candidates clearly did not read the stem carefully as they referred to the mass being on an incline.

2.

[Maximum mark: 12]

22M.2.SL.TZ1.2

Cold milk enters a small sterilizing unit and flows over an electrical heating element.

The temperature of the milk is raised from 11 °C to 84 °C. A mass of 55 g of milk enters the sterilizing unit every second.

Specific heat capacity of milk = 3.9 kJ kg⁻¹K⁻¹

The milk flows out through an insulated metal pipe. The pipe is at a temperature of 84 °C. A small section of the insulation has been removed from around the pipe.

(a)

Estimate the power input to the heating element. State an appropriate unit for your answer.

[2]

Markscheme

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Examiners report

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

(b)

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

[2]

Markscheme

Underestimate / more energy or power required ✓

because energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓

Do not allow general term “energy” or “power” for MP2.

Examiners report

Many candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as "some power will be lost"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).

(c)

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

[2]

Markscheme

the temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓

«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓

Examiners report

This was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.

(d.i)

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

[1]

Markscheme

conduction/conducting/conductor «through metal» ✓

Examiners report

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

(d.ii)

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

[3]

Markscheme

use of $P = e σ A T^{4}$ where T = 357 K ✓

use of $A = 2 π r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

Examiners report

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

(d.iii)

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

[2]

Markscheme

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Examiners report

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

a.

[Maximum mark: 2]

22M.2.SL.TZ1.2a

(a)

Estimate the power input to the heating element. State an appropriate unit for your answer.

[2]

Markscheme

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Examiners report

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

b.

[Maximum mark: 2]

22M.2.SL.TZ1.2b

(b)

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

[2]

Markscheme

Underestimate / more energy or power required ✓

because energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓

Do not allow general term “energy” or “power” for MP2.

Examiners report

Many candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as "some power will be lost"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).

c.

[Maximum mark: 2]

22M.2.SL.TZ1.2c

(c)

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

[2]

Markscheme

the temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓

«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓

Examiners report

This was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.

d.i.

[Maximum mark: 1]

22M.2.SL.TZ1.2d.i

(d.i)

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

[1]

Markscheme

conduction/conducting/conductor «through metal» ✓

Examiners report

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

d.ii.

[Maximum mark: 3]

22M.2.SL.TZ1.2d.ii

(d.ii)

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

[3]

Markscheme

use of $P = e σ A T^{4}$ where T = 357 K ✓

use of $A = 2 π r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

Examiners report

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

d.iii.

[Maximum mark: 2]

22M.2.SL.TZ1.2d.iii

(d.iii)

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

[2]

Markscheme

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Examiners report

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

3.

[Maximum mark: 7]

22M.2.SL.TZ1.3

Two loudspeakers A and B are initially equidistant from a microphone M. The frequency and intensity emitted by A and B are the same. A and B emit sound in phase. A is fixed in position.

B is moved slowly away from M along the line MP. The graph shows the variation with distance travelled by B of the received intensity at M.

(a)

Explain why the received intensity varies between maximum and minimum values.

[3]

Markscheme

movement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓

interference
OR
superposition «of waves» ✓

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or $π$ » out of phase / path difference = (n+½) x lambda ✓

Examiners report

This was an "explain" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. "there is constructive and destructive interference"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.

(b)

State and explain the wavelength of the sound measured at M.

[2]

Markscheme

wavelength = 26 cm ✓

peak to peak distance is the path difference which is one wavelength

OR

this is the distance B moves to be back in phase «with A» ✓

Allow 25 − 27 cm for MP1.

Examiners report

Many candidates recognized that the wavelength was 26 cm, but the explanations were lacking the details about what information the graph was actually providing. Examiners were looking for a connection back to path difference, and not simply a description of peak-to-peak distance on the graph. Some candidates did not state a wavelength at all, and instead simply discussed the concept of wavelength or suggested that the wavelength was constant.

(c)

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

Speed of sound = 340 m s⁻¹

[2]

Markscheme

« $\frac{λ}{2}$ » = 13 cm ✓

$f =$ « $\frac{c}{λ} = \frac{340}{0.13} =$ » 2.6 «kHz» ✓

Allow ½ of wavelength from (b) or data from graph.

Examiners report

This was a "show that" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.

a.

[Maximum mark: 3]

22M.2.SL.TZ1.3a

(a)

Explain why the received intensity varies between maximum and minimum values.

[3]

Markscheme

movement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓

interference
OR
superposition «of waves» ✓

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or $π$ » out of phase / path difference = (n+½) x lambda ✓

Examiners report

This was an "explain" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. "there is constructive and destructive interference"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.

b.

[Maximum mark: 2]

22M.2.SL.TZ1.3b

(b)

State and explain the wavelength of the sound measured at M.

[2]

Markscheme

wavelength = 26 cm ✓

peak to peak distance is the path difference which is one wavelength

OR

this is the distance B moves to be back in phase «with A» ✓

Allow 25 – 27 cm for MP1.

Examiners report

Many candidates recognized that the wavelength was 26 cm, but the explanations were lacking the details about what information the graph was actually providing. Examiners were looking for a connection back to path difference, and not simply a description of peak-to-peak distance on the graph. Some candidates did not state a wavelength at all, and instead simply discussed the concept of wavelength or suggested that the wavelength was constant.

c.

[Maximum mark: 2]

22M.2.SL.TZ1.3c

(c)

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

Speed of sound = 340 m s⁻¹

[2]

Markscheme

« $\frac{λ}{2}$ » = 13 cm ✓

$f =$ « $\frac{c}{λ} = \frac{340}{0.13} =$ » 2.6 «kHz» ✓

Allow ½ of wavelength from (b) or data from graph for MP1.

Allow ECF from MP1.

Examiners report

This was a "show that" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.

a.

[Maximum mark: 2]

22M.2.SL.TZ1.6a

(a)

Outline two reasons why both models predict that the motion is simple harmonic when $a$ is small.

[2]

Markscheme

For both models:
displacement is ∝ to acceleration/force «because graph is straight and through origin» ✓

displacement and acceleration / force in opposite directions «because gradient is negative»
OR
acceleration/«restoring» force is always directed to equilibrium ✓

Examiners report

This item was essentially encouraging candidates to connect concepts about simple harmonic motion to a physical situation described by a graph. The marks were awarded for discussing the physical motion (such as "the acceleration is in the opposite direction of the displacement") and not just for describing the graph itself (such as "the slope of the graph is negative"). Most candidates were successful in recognizing that the acceleration was proportional to displacement for the first marking point, but many simply described the graph for the second marking point.

b.

[Maximum mark: 4]

22M.2.SL.TZ1.6b

(b)

Determine the time period of the system when $a$ is small.

[4]

Markscheme

attempted use of $ω^{2} = (-) \frac{a}{x}$ ✓

suitable read-offs leading to gradient of line = 28 « s^-2» ✓

$T = \frac{2 π}{ω}$ « $\frac{2 π}{\sqrt{28}}$ » ✓

$T = 1.2$ s ✓

Examiners report

This question was well done by many candidates. A common mistake was to select an incorrect gradient, but candidates who showed their work clearly still earned the majority of the marks.

c.

[Maximum mark: 2]

22M.2.SL.TZ1.6c

(c)

Outline, without calculation, the change to the time period of the system for the model represented by graph B when $a$ is large.

[2]

Markscheme

time period increases ✓

because average ω «for whole cycle» is smaller

OR

slope / acceleration / force at large x is smaller

OR

area under graph B is smaller so average speed is smaller. ✓

Examiners report

Many candidates recognized that the time period would increase for B, and some were able to give a valid reason based on the difference between the motion of B and the motion of A. It should be noted that the prompt specified "without calculation", so candidates who simply attempted to calculate the time period of B did not receive marks.

d.

[Maximum mark: 2]

22M.2.SL.TZ1.6d

(d)

The graph shows for model A the variation with $x$ of elastic potential energy E_p stored in the spring.

Describe the graph for model B.

[2]

Markscheme

same curve OR shape for small amplitudes «to about 0.05 m» ✓

for large amplitudes «outside of 0.05 m» E_p smaller for model B / values are lower than original / spread will be wider ✓ OWTTE

Accept answers drawn on graph – e.g.

Examiners report

Candidates were generally successful in describing one of the two aspects of the graph of B compared to A, but few were able to describe both. It should be noted that this is a two mark question, so candidates should have considered the fact that there are two distinct statements to be made about the graphs. Examiners did accept clearly drawn graphs as well for full marks.

a.

[Maximum mark: 2]

22M.2.SL.TZ1.a

Estimate the power input to the heating element. State an appropriate unit for your answer.

[2]

Markscheme

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Examiners report

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

b.

[Maximum mark: 4]

22M.2.SL.TZ1.b

Determine the time period of the system when $a$ is small.

[4]

Markscheme

attempted use of $ω^{2} = (-) \frac{a}{x}$ ✓

suitable read-offs leading to gradient of line = 28 « s^-2» ✓

$T = \frac{2 π}{ω}$ « $\frac{2 π}{\sqrt{28}}$ » ✓

$T = 1.2$ s ✓

Examiners report

This question was well done by many candidates. A common mistake was to select an incorrect gradient, but candidates who showed their work clearly still earned the majority of the marks.

c.

[Maximum mark: 2]

22M.2.SL.TZ1.c

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

Speed of sound = 340 m s⁻¹

[2]

Markscheme

« $\frac{λ}{2}$ » = 13 cm ✓

$f =$ « $\frac{c}{λ} = \frac{340}{0.13} =$ » 2.6 «kHz» ✓

Allow ½ of wavelength from (b) or data from graph for MP1.

Allow ECF from MP1.

Examiners report

This was a "show that" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.

c.i.

[Maximum mark: 2]

22M.2.SL.TZ1.c.i

Show that the speed of the load when it hits the floor is about 2.1 m s⁻¹.

[2]

Markscheme

proper use of kinematic equation ✓

$\sqrt{(2 \times 2.4 \times 0.95)} = 2.14$ «m s⁻¹» ✓

Must see either the substituted values OR a value for v to at least three s.f. for MP2.

Examiners report

This was a "show that" questions, so examiners were looking for a clear equation leading to a clear substitution of values leading to an answer that had more significant digits than the given answer. Most candidates successfully selected the correct equation and showed a proper substitution. Some candidates started with an energy approach that needed modification as it clearly led to an incorrect solution. These responses did not receive full marks.

c.ii.

[Maximum mark: 2]

22M.2.SL.TZ1.c.ii

The radius of the pulley is 2.5 cm. Calculate the angular speed of rotation of the pulley as the load hits the floor. State your answer to an appropriate number of significant figures.

[2]

Markscheme

use of $ω = \frac{v}{r}$ to give 84 «rad s⁻¹»

OR

$ω = 2.1 / 0.025$ to give 84 «rad s⁻¹» ✓

quoted to 2sf only✓

Examiners report

This SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.

d.

[Maximum mark: 4]

22M.2.SL.TZ1.d

After the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.

[4]

Markscheme

ALTERNATIVE 1

« $v^{2} = u^{2} + 2 a s \Rightarrow 0 = 2 . 1^{2} - 2 a \times 0.35$ » leading to a = 6.3 «m s^-2»

OR

« $x = 1 / 2 (u + v) t$ » leading to t = 0.33 « s » ✓

F_net = « $m a = 1.5 \times 6.3$ = » 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

ALTERNATIVE 2

kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)² = F_NET x 0.35 ✓

F_net = 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

Accept 1.95 N from g = 10 m s^-2.
Accept 2.42 N from u = 2.14 m s^-1.

Examiners report

Candidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.

d.i.

[Maximum mark: 1]

22M.2.SL.TZ1.d.i

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

[1]

Markscheme

conduction/conducting/conductor «through metal» ✓

Examiners report

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

d.ii.

[Maximum mark: 3]

22M.2.SL.TZ1.d.ii

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

[3]

Markscheme

use of $P = e σ A T^{4}$ where T = 357 K ✓

use of $A = 2 π r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

Examiners report

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

d.iii.

[Maximum mark: 2]

22M.2.SL.TZ1.d.iii

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

[2]

Markscheme

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Examiners report

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

e.

[Maximum mark: 3]

22M.2.SL.TZ1.e

The student then makes the ramp horizontal and applies a constant horizontal force to the box. The force is just large enough to start the box moving. The force continues to be applied after the box begins to move.

Explain, with reference to the frictional force acting, why the box accelerates once it has started to move.

[3]

Markscheme

static coefficient of friction > dynamic/kinetic coefficient of friction / μ_s > μ_k ✓

«therefore» force of dynamic/kinetic friction will be less than the force of static friction ✓

there will be a net / unbalanced forward force once in motion «which results in acceleration»

OR

reference to net F = ma ✓

Examiners report

This was an "explain" question, so examiners were looking for a clear line of discussion starting with a comparison of the coefficients of friction, leading to a comparison of the relative magnitudes of the forces of friction and ultimately the rise of a net force leading to an acceleration. Many candidates recognized that this was a question about the comparison between static and kinetic/dynamic friction but did not clearly specify which they were referring to in their responses. Some candidates clearly did not read the stem carefully as they referred to the mass being on an incline.

20.

[Maximum mark: 1]

22N.1A.SL.TZ0.20

A car on a road follows a horizontal circular path at a constant speed. What is the direction of the net force acting on the car and the direction of the instantaneous velocity of the car?

[1]

Markscheme

C

3.

[Maximum mark: 1]

22N.1A.SL.TZ0.3

Ball 1 is dropped from rest from an initial height $h$ . At the same instant, ball 2 is launched vertically upwards at an initial velocity $u$ .

At what time are both balls at the same distance above the ground?

A. $\frac{h}{4 u}$

B. $\frac{h}{2 u}$

C. $\frac{h}{u}$

D. $\frac{2 h}{u}$

[1]

Markscheme

C

Examiners report

This question turned out to be challenging especially to SL candidates, and the number of blank answers was higher than expected. The most efficient way of solving it is by equating algebraic expressions for the height of each ball as a function of time, e.g. $h - \frac{1}{2} g t^{2} = u t - \frac{1}{2} g t^{2}$ , followed by cancellation of the quadratic term.

30.

[Maximum mark: 1]

22N.1A.SL.TZ0.30

Planet $X$ and planet $Y$ both emit radiation as black bodies. Planet $Y$ has twice the surface temperature and one third of the radius of planet $X$ .

What is $\frac{power radiated by planet X}{power radiated by planet Y}$ ?

A. $\frac{9}{16}$

B. $\frac{3}{4}$

C. $\frac{4}{3}$

D. $\frac{16}{9}$

[1]

Markscheme

A

4.

[Maximum mark: 1]

22N.1A.SL.TZ0.4

A projectile is launched with a velocity $u$ at an angle $θ$ to the horizontal. It reaches a maximum height $s$ . What is the time taken to reach the maximum height?

A. $\frac{2 s}{u \cos θ}$

B. $\frac{2 s}{g}$

C. $\frac{u \cos θ}{g}$

D. $\frac{u \sin θ}{g}$

[1]

Markscheme

D

Examiners report

This question is very easily answered by elimination. The time to reach max height is the same as for a projectile launched vertically with initial speed $u \sin θ$ , and no other alternative than D contains the term $u \sin θ$ .

5.

[Maximum mark: 1]

22N.1A.SL.TZ0.5

The diagram shows the trajectory of a projectile and the velocity v of the projectile at point P in its trajectory. P is located before the projectile reaches the peak altitude. Air resistance acts on the projectile. The acceleration of the projectile at P is a.

What are the magnitudes of the horizontal component and the vertical component of the acceleration of the projectile at P?

[1]

Markscheme

B

1.

[Maximum mark: 8]

22N.2.SL.TZ0.1

A raindrop falls vertically from rest.

The graph shows how the speed of the raindrop varies with time t.

During the first 3.0 s of motion, the raindrop falls a distance of 21 m and reaches a speed of 9.0 m s⁻¹. The mass of the raindrop is 34 mg. The temperature of the raindrop does not change.

(a)

State the initial acceleration of the raindrop.

[1]

Markscheme

g OR 9.81 «m s⁻²» OR acceleration of gravity/due to free fall ✓

Accept 10 «m s⁻²».

Ignore sign.

Do not accept bald “gravity”.

Accept answer that indicates tangent of the graph at time t=0.

Examiners report

A nice introductory question answered correctly by most candidates. Most answers quoted the data booklet value, with a few 10's or 9.8's, or the answer in words. Very few lost the mark by just stating gravity, or zero.

(b)

Explain, by reference to the vertical forces, how the raindrop reaches a constant speed.

[3]

Markscheme

Identification of air resistance/drag force «acting upwards» ✓

«that» increases with speed ✓

«until» weight and air resistance cancel out

OR

net force/acceleration becomes zero ✓

A statement as “air resistance increases with speed” scores MP1 and MP2.

Examiners report

This was very well answered with most candidates scoring 3. The MP usually missed in candidates scoring 2 marks was MP2, to justify the variation of the magnitude of air resistance, although that rarely happened.

(c.i)

Determine the energy transferred to the air during the first 3.0 s of motion. State your answer to an appropriate number of significant figures.

[3]

Markscheme

«loss in» GPE = 3.4 × 10⁻⁵ × 9.81 × 21 «= 7.0 × 10⁻³» «J»

OR

«gain in» KE = 0.5 × 3.4 × 10⁻⁵ × 9.0² «= 1.4 × 10⁻³» «J» ✓

energy transferred to air «=7.0 × 10⁻³ − 1.4 × 10⁻³» = 5.6 × 10⁻³» «J» ✓

any calculated answer to 2 sf ✓

Allow [1] through the use of kinematics assuming constant acceleration.

Allow ECF from MP1.

Examiners report

Generally well answered, although several candidates lost a mark, usually as POT (power of ten) by quoting the value in kg leading to an answer of 5.3 J. Most candidates were able to score MP3 by rounding their calculation to two significant figures.

(c.ii)

Describe the energy change that takes place for t > 3.0 s.

[1]

Markscheme

«gravitational» potential energy «of the raindrop» into thermal/internal energy «of the air» ✓

Accept heat for thermal energy.

Accept into kinetic energy of air particles.

Ignore sound energy.

Examiners report

Of the wrong answers, the most common ones were gravitational potential to kinetic or the idea that because there was no change in velocity there was no energy transfer. A significant number, though, scored by identifying the change into thermal (most of them), kinetic of air particles (a few answers) or internal (very few).

a.

[Maximum mark: 1]

22N.2.SL.TZ0.1a

(a)

State the initial acceleration of the raindrop.

[1]

Markscheme

g OR 9.81 «m s⁻²» OR acceleration of gravity/due to free fall ✓

Accept 10 «m s⁻²».

Ignore sign.

Do not accept bald “gravity”.

Accept answer that indicates tangent of the graph at time t=0.

Examiners report

A nice introductory question answered correctly by most candidates. Most answers quoted the data booklet value, with a few 10's or 9.8's, or the answer in words. Very few lost the mark by just stating gravity, or zero.

b.

[Maximum mark: 3]

22N.2.SL.TZ0.1b

(b)

Explain, by reference to the vertical forces, how the raindrop reaches a constant speed.

[3]

Markscheme

Identification of air resistance/drag force «acting upwards» ✓

«that» increases with speed ✓

«until» weight and air resistance cancel out

OR

net force/acceleration becomes zero ✓

A statement as “air resistance increases with speed” scores MP1 and MP2.

Examiners report

This was very well answered with most candidates scoring 3. The MP usually missed in candidates scoring 2 marks was MP2, to justify the variation of the magnitude of air resistance, although that rarely happened.

c.i.

[Maximum mark: 3]

22N.2.SL.TZ0.1c.i

(c.i)

Determine the energy transferred to the air during the first 3.0 s of motion. State your answer to an appropriate number of significant figures.

[3]

Markscheme

«loss in» GPE = 3.4 × 10⁻⁵ × 9.81 × 21 «= 7.0 × 10⁻³» «J»

OR

«gain in» KE = 0.5 × 3.4 × 10⁻⁵ × 9.0² «= 1.4 × 10⁻³» «J» ✓

energy transferred to air «=7.0 × 10⁻³ − 1.4 × 10⁻³» = 5.6 × 10⁻³» «J» ✓

any calculated answer to 2 sf ✓

Allow [1] through the use of kinematics assuming constant acceleration.

Allow ECF from MP1.

Examiners report

Generally well answered, although several candidates lost a mark, usually as POT (power of ten) by quoting the value in kg leading to an answer of 5.3 J. Most candidates were able to score MP3 by rounding their calculation to two significant figures.

c.ii.

[Maximum mark: 1]

22N.2.SL.TZ0.1c.ii

(c.ii)

Describe the energy change that takes place for t > 3.0 s.

[1]

Markscheme

«gravitational» potential energy «of the raindrop» into thermal/internal energy «of the air» ✓

Accept heat for thermal energy.

Accept into kinetic energy of air particles.

Ignore sound energy.

Examiners report

Of the wrong answers, the most common ones were gravitational potential to kinetic or the idea that because there was no change in velocity there was no energy transfer. A significant number, though, scored by identifying the change into thermal (most of them), kinetic of air particles (a few answers) or internal (very few).

a.i.

[Maximum mark: 3]

22N.2.SL.TZ0.2a.i

(a.i)

Determine the minimum area of the solar heating panel required to increase the temperature of all the water in the tank to 30°C during a time of 1.0 hour.

[3]

Markscheme

energy required = 250 × 4200 × (30 − 15) ✓

energy available = 0.30 × 680 × t × A ✓

A = « $\frac{250 \times 4200 \times 15}{0.30 \times 680 \times 60 \times 60} =$ » 21 «m²» OR 22 «m²» ✓

Allow ECF from MP1 and MP2.

Accept the correct use of 0.30 in either MP1 or MP2.

Examiners report

Most candidates had a good attempt at this but there were often slight slips. Some missed the efficiency of the process. Some included the albedo of the roof tiles. Some thought that the temperature rise needed to have 273 added to convert to kelvin. However, sometimes scoring through ECF (error carried forward), the average mark was around 2 marks.

a.ii.

[Maximum mark: 3]

22N.2.SL.TZ0.2a.ii

(a.ii)

Estimate, in °C, the temperature of the roof tiles.

[3]

Markscheme

absorbed intensity = (1 − 0.2) × 680 «= 544» «W m⁻²»

OR

emitted intensity = 0.97 × 5.67 × 10⁻⁸ × T⁴ ✓

T $= \sqrt[4]{\frac{544}{0.97 \times 5.67 \times 10^{- 8}}} = 315$ «K» ✓

42 «°C» ✓

Allow ECF from MP1 and MP2.

Allow MP1 if absorbed or emitted intensity is multiplied by area.

Examiners report

This was a bit more hit and miss than the previous question part. One common mistake was not understanding what albedo meant. Some took it as the amount of energy absorbed rather than reflected. Emissivity was often missed. Several candidates, successfully answering the question or not, were able to score MP3 converting the final temperature into Celsius degrees.

b.i.

[Maximum mark: 1]

22N.2.SL.TZ0.2b.i

(b.i)

State one way in which a real gas differs from an ideal gas.

[1]

Markscheme

can be liquefied ✓

has intermolecular forces / potential energy ✓

has atoms/molecules that are not point objects / take up volume ✓

does not follow the ideal gas law «for all T and p» ✓

collisions between particles are non-elastic ✓

Accept the converse argument.

Examiners report

This was very well answered. Candidates showed an understanding of the differences between ideal and real gases.

b.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.2b.ii

(b.ii)

The water is heated. Explain why the quantity of air in the storage tank decreases.

[2]

Markscheme

ALTERNATIVE 1

«constant p and V imply» nT = const ✓

T increases hence n decreases ✓

ALTERNATIVE 2

«constant p and n imply» V is proportional to T / air expands as it is heated ✓

«original» air occupies a greater volume OR some air leaves through opening ✓

MP2 in ALT 2 must come from expansion of air, not from expansion of water.

Award [0] for an answer based on expansion of water.

Award [1] max for an answer based on convection currents.

Examiners report

It was surprising to see a large number of answers based on the expansion of water, as the stem of the question clearly states that the level of water remains constant. Most successful candidates scored by quoting pV constant so concluding with the inverse relationship of n and T, others also managed to score by explaining that the volume of air increases and therefore must go out through the opening. Answers based on convection currents were given partial credit.

3.

[Maximum mark: 10]

22N.2.SL.TZ0.3

A string of length 0.80 m is fixed at both ends. The diagram shows a standing wave formed on the string. P and Q are two particles on the string.

The variation with time t of the displacement of particle P is shown.

It is suggested that the speed c of waves in the string is related to the tension force T in the string according to the equation T = ac², where a is a constant.

(a.i)

Draw, on the axes, a graph to show the variation with t of the displacement of particle Q.

[2]

Markscheme

oscillation in antiphase ✓

smaller amplitude than P ✓

Examiners report

Although there were good answers which scored full marks, there were a significant number of wrong answers where the amplitude was the same or not consistent throughout, or the wave drawn was not in antiphase of the original sketch.

(a.ii)

Calculate the speed of waves on the string.

[2]

Markscheme

wavelength $= \frac{2}{3} \times 0.80 = 0.53$ «m» ✓

speed $= \frac{0.53}{2.8 \times 10^{- 3}} = 190$ «m s⁻¹» ✓

Allow ECF from incorrect wavelength.

Examiners report

This was well answered, particularly MP1 to determine the wavelength, although several candidates misinterpreted the unit of time and obtained a very small value for the velocity of the wave.

(b.i)

Determine the fundamental SI unit for a.

[2]

Markscheme

kg m s⁻² OR m² s⁻² seen ✓

kg m⁻¹ ✓

Award [2] for a BCA.

Examiners report

Students seem to be well prepared for this sort of question, as it was high-scoring.

(b.ii)

The tension force on the string is doubled. Describe the effect, if any, of this change on the frequency of the standing wave.

[2]

Markscheme

speed increases hence frequency increases ✓

by factor $\sqrt{2}$ ✓

Examiners report

This question was answered well, although the numerical aspect was often missing. It is worth highlighting that if there is a term like 'doubled' in the question, it makes sense to expect a numerical answer.

(c)

The standing wave on the string creates a travelling sound wave in the surrounding air.

Outline two differences between a standing wave and a travelling wave.

[2]

Markscheme

travelling waves transfer energy OR standing waves don’t ✓

amplitude of oscillation varies along a standing wave OR is constant along a travelling wave ✓

standing waves have nodes and antinodes OR travelling waves don’t ✓

points in an internodal region have same phase in standing waves OR different phase in travelling waves ✓

Examiners report

This question was answered well. Students showed to be familiar with the differences between standing and travelling waves. In SL they had to identify two differences, so that proved to be more challenging.

a.i.

[Maximum mark: 2]

22N.2.SL.TZ0.3a.i

(a.i)

Draw, on the axes, a graph to show the variation with t of the displacement of particle Q.

[2]

Markscheme

oscillation in antiphase ✓

smaller amplitude than P ✓

Examiners report

Although there were good answers which scored full marks, there were a significant number of wrong answers where the amplitude was the same or not consistent throughout, or the wave drawn was not in antiphase of the original sketch.

a.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.3a.ii

(a.ii)

Calculate the speed of waves on the string.

[2]

Markscheme

wavelength $= \frac{2}{3} \times 0.80 = 0.53$ «m» ✓

speed $= \frac{0.53}{2.8 \times 10^{- 3}} = 190$ «m s⁻¹» ✓

Allow ECF from incorrect wavelength.

Examiners report

This was well answered, particularly MP1 to determine the wavelength, although several candidates misinterpreted the unit of time and obtained a very small value for the velocity of the wave.

b.i.

[Maximum mark: 2]

22N.2.SL.TZ0.3b.i

(b.i)

Determine the fundamental SI unit for a.

[2]

Markscheme

kg m s⁻² OR m² s⁻² seen ✓

kg m⁻¹ ✓

Award [2] for a BCA.

Examiners report

Students seem to be well prepared for this sort of question, as it was high-scoring.

b.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.3b.ii

(b.ii)

The tension force on the string is doubled. Describe the effect, if any, of this change on the frequency of the standing wave.

[2]

Markscheme

speed increases hence frequency increases ✓

by factor $\sqrt{2}$ ✓

Examiners report

This question was answered well, although the numerical aspect was often missing. It is worth highlighting that if there is a term like 'doubled' in the question, it makes sense to expect a numerical answer.

c.

[Maximum mark: 2]

22N.2.SL.TZ0.3c

(c)

The standing wave on the string creates a travelling sound wave in the surrounding air.

Outline two differences between a standing wave and a travelling wave.

[2]

Markscheme

travelling waves transfer energy OR standing waves don’t ✓

amplitude of oscillation varies along a standing wave OR is constant along a travelling wave ✓

standing waves have nodes and antinodes OR travelling waves don’t ✓

points in an internodal region have same phase in standing waves OR different phase in travelling waves ✓

Examiners report

This question was answered well. Students showed to be familiar with the differences between standing and travelling waves. In SL they had to identify two differences, so that proved to be more challenging.

c.i.

[Maximum mark: 1]

22N.2.SL.TZ0.3c.i

(c.i)

Outline one difference between a standing wave and a travelling wave.

[1]

Markscheme

travelling waves transfer energy OR standing waves don’t ✓

amplitude of oscillation varies along a standing wave OR is constant along a travelling wave ✓

standing waves have nodes / antinodes OR travelling waves don’t ✓

points in an internodal region have same phase in standing waves OR different phase in travelling waves ✓

Examiners report

This question was answered well. Students showed to be familiar with the differences between standing and travelling waves.

c.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.3c.ii

(c.ii)

The speed of sound in air is 340 m s⁻¹ and in water it is 1500 m s⁻¹.

Discuss whether the sound wave can enter the water.

[2]

Markscheme

ALTERNATIVE 1

critical angle $= 13 °$ «from $\sin θ_{c} = \frac{340}{1500}$ » ✓

the angle of incidence is greater than $θ_{c}$ hence the sound can’t enter water ✓

ALTERNATIVE 2

$\sin θ_{r} = \frac{1500}{340} \sin 30 °$ ✓

sine value greater than one hence the sound can’t enter water ✓

Conclusion must be justified, award [0] for BCA.

Examiners report

Surprisingly well answered as it was sound from air to water, rather than light from air to glass. A mixture of approaches but probably the most common was to calculate a sine value of over 1. Some went about calculating the critical angle but nowhere near as many.

4.

[Maximum mark: 5]

22N.2.SL.TZ0.4

A mass is attached to one end of a rod and made to rotate with constant speed in a vertical circle.

(a)

The scale diagram shows the weight W of the mass at an instant when the rod is horizontal.

Draw, on the scale diagram, an arrow to represent the force exerted on the mass by the rod.

[2]

Markscheme

horizontal component of any length to the left ✓

vertical component two squares long upwards ✓

E.g.

Ignore point of application.

Award [1] max if arrowhead not present.

Examiners report

Most just added the horizontal component. Not many centrifugal forces, but still a few. Very few were able to score both marks, so this question proved to be challenging for the candidates.

(b)

Explain why the magnitude of the force exerted on the mass by the rod is not constant.

[3]

Markscheme

ALTERNATIVE 1

the net/centripetal force has constant magnitude ✓

the direction of the net/centripetal force constantly changes ✓

this is achieved by vector-adding weight and the force from the rod

OR

the force from the rod is vector difference of the centripetal force and weight ✓

ALTERNATIVE 2

at the top F_rod = F_c − W ✓

at the bottom, F_rod = F_c + W ✓

net F/F_c is constant so the force from the rod is different «hence is changing» ✓

Accept reference to centripetal or net force indistinctly.

Allow reference to centripetal acceleration.

Examiners report

Many got into a bit of a mess with this one and it was quite difficult to interpret some of the answers. If they started out with the net/centripetal force being constant, then it was often easy to follow the reasoning. Starting with force on the rod varying often led to confusion. Quite a few did not pick up on the constant speed vertical circle so there were complicated energy/speed arguments to pick through.

a.

[Maximum mark: 2]

22N.2.SL.TZ0.4a

(a)

The scale diagram shows the weight W of the mass at an instant when the rod is horizontal.

Draw, on the scale diagram, an arrow to represent the force exerted on the mass by the rod.

[2]

Markscheme

horizontal component of any length to the left ✓

vertical component two squares long upwards ✓

E.g.

Ignore point of application.

Award [1] max if arrowhead not present.

Examiners report

Most just added the horizontal component. Not many centrifugal forces, but still a few. Very few were able to score both marks, so this question proved to be challenging for the candidates.

b.

[Maximum mark: 3]

22N.2.SL.TZ0.4b

(b)

Explain why the magnitude of the force exerted on the mass by the rod is not constant.

[3]

Markscheme

ALTERNATIVE 1

the net/centripetal force has constant magnitude ✓

the direction of the net/centripetal force constantly changes ✓

this is achieved by vector-adding weight and the force from the rod

OR

the force from the rod is vector difference of the centripetal force and weight ✓

ALTERNATIVE 2

at the top F_rod = F_c − W ✓

at the bottom, F_rod = F_c + W ✓

net F/F_c is constant so the force from the rod is different «hence is changing» ✓

Accept reference to centripetal or net force indistinctly.

Allow reference to centripetal acceleration.

Examiners report

Many got into a bit of a mess with this one and it was quite difficult to interpret some of the answers. If they started out with the net/centripetal force being constant, then it was often easy to follow the reasoning. Starting with force on the rod varying often led to confusion. Quite a few did not pick up on the constant speed vertical circle so there were complicated energy/speed arguments to pick through.

5.

[Maximum mark: 8]

22N.2.SL.TZ0.5

Resistor R is connected in a circuit with a cell that has internal resistance.

The ammeter and the voltmeter are ideal.

The cell has an emf of 1.49 V. The resistance of R is 50.0 Ω. The voltmeter reads 1.47 V.

One of the connecting wires is placed in a magnetic field. The direction of the current in the wire is shown.

(a)

State what is meant by an ideal voltmeter.

[1]

Markscheme

infinite resistance

OR

no current is flowing through it ✓

Examiners report

A majority of candidates scored a mark by simply stating infinite resistance. Several answers went the other way round, stating a resistance of zero.

(b.i)

Show that the internal resistance of the cell is about 0.7 Ω.

[2]

Markscheme

current $= \frac{1.47}{50.0} = 2.94 \times 10^{- 2}$ «A» ✓

r $= \frac{1.49}{2.94 \times 10^{- 2}} - 50.0$ OR $0.68$ «Ω» ✓

For MP2, allow any other correctly substituted expression for r.

Examiners report

Many answers here produced a number that did not round to 0.7 but students claimed it did. The simultaneous equation approach was seen in the best candidates, getting the right answer. It is worthy of reminding about the need of showing one more decimal place when calculating a show that value type of question.

(b.ii)

Determine the total power dissipated in the circuit.

[2]

Markscheme

$1.49 \times 2.94 \times 10^{- 2}$

OR

$\frac{1 . 49^{2}}{50.0 + 0.68}$

OR

${(2.94 \times 10^{- 2})}^{2} (50.0 + 0.68)$ ✓

$0.0438$ «W» ✓

Accept use of 0.7 Ω in MP1.

Examiners report

The calculation of total power dissipated was not usually well done, as it often failed to include the internal resistance therefore calculating the power dissipated only in the resistor.

(c.i)

Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.

[2]

Markscheme

charge/carriers are moving in a magnetic field ✓

there is a magnetic force on them / quote F = qvB

OR

this creates a magnetic field that interacts with the external magnetic field ✓

Accept electrons.

For MP2, the force must be identified as acting on charge / carriers.

Examiners report

Many scored MP1 here but did not get MP2 as they jumped straight to the wire rather than continuing with the explanation of what was going on with the charge carriers.

(c.ii)

Every current-carrying wire produces a magnetic field.

Describe one piece of evidence that supports this statement.

[1]

Markscheme

magnetic needle is deflected by nearby currents

OR

two «parallel» current-carrying wires exert a force on each other

OR

magnetic field due to a current can be measured directly with a probe ✓

Only accept argument that refers to an observation or experiment.

Examiners report

Students usually failed to identify appropriate evidence.

a.

[Maximum mark: 1]

22N.2.SL.TZ0.5a

(a)

State what is meant by an ideal voltmeter.

[1]

Markscheme

infinite resistance

OR

no current is flowing through it ✓

Examiners report

A majority of candidates scored a mark by simply stating infinite resistance. Several answers went the other way round, stating a resistance of zero.

b.i.

[Maximum mark: 1]

22N.2.SL.TZ0.5b.i

(b.i)

Calculate, in mA, the current in the resistor.

[1]

Markscheme

$\frac{1.47}{50.0} = 29.4$ «mA» ✓

Examiners report

Very well answered, with most candidates successfully answering in the required unit of mA.

b.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.5b.ii

(b.ii)

Determine the total power dissipated in the circuit.

[2]

Markscheme

$1.49 \times 2.94 \times 10^{- 2}$

OR

$\frac{1 . 49^{2}}{50.0 + 0.68}$

OR

${(2.94 \times 10^{- 2})}^{2} (50.0 + 0.68)$ ✓

$0.0438$ «W» ✓

Accept use of 0.7 Ω in MP1.

Examiners report

The calculation of total power dissipated was not usually well done, as it often failed to include the internal resistance therefore calculating the power dissipated only in the resistor.

b.iii.

[Maximum mark: 2]

22N.2.SL.TZ0.5b.iii

(b.iii)

Calculate the emf of the cell.

[2]

Markscheme

139 × 10⁻³(10.0 + 0.73)

OR

29.4 × 10⁻³ (50.0 + 0.73) ✓

1.49 «V» ✓

Watch for ECF from 5(b)(i).

Examiners report

Usually well answered, regardless of b(ii), by utilising the show that value given.

c.i.

[Maximum mark: 2]

22N.2.SL.TZ0.5c.i

(c.i)

Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.

[2]

Markscheme

charge/carriers are moving in a magnetic field ✓

there is a magnetic force on them / quote F = qvB

OR

this creates a magnetic field that interacts with the external magnetic field ✓

Accept electrons.

For MP2, the force must be identified as acting on charge / carriers.

Examiners report

Many scored MP1 here but did not get MP2 as they jumped straight to the wire rather than continuing with the explanation of what was going on with the charge carriers.

c.ii.

[Maximum mark: 1]

22N.2.SL.TZ0.5c.ii

(c.ii)

Every current-carrying wire produces a magnetic field.

Describe one piece of evidence that supports this statement.

[1]

Markscheme

magnetic needle is deflected by nearby currents

OR

two «parallel» current-carrying wires exert a force on each other

OR

magnetic field due to a current can be measured directly with a probe ✓

Only accept argument that refers to an observation or experiment.

Examiners report

Students usually failed to identify appropriate evidence.

6.

[Maximum mark: 9]

22N.2.SL.TZ0.6

Polonium-210 (Po-210) decays by alpha emission into lead-206 (Pb-206).

The following data are available.

Nuclear mass of Po-210 = 209.93676 u

Nuclear mass of Pb-206 = 205.92945 u

Mass of the alpha particle = 4.00151 u

(a)

Outline, by reference to nuclear binding energy, why the mass of a nucleus is less than the sum of the masses of its constituent nucleons.

[2]

Markscheme

according to ΔE = Δmc² / identifies mass energy equivalence ✓

energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»

OR

energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓

Accept protons and neutrons.

Examiners report

Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.

(b.i)

Calculate, in MeV, the energy released in this decay.

[2]

Markscheme

(m_polonium− m_lead − m_α)c² OR (209.93676 − 205.92945 − 4.00151)

OR

mass difference = 5.8 × 10⁻³ ✓

conversion to MeV using 931.5 to give 5.4 «MeV» ✓

Allow ECF from MP1.

Award [2] for a BCA.

Award [1] for 8.6 x 10⁻¹³ J.

Examiners report

Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c² as they were not sure of the significance of the unit of u.

(b.ii)

The polonium nucleus was stationary before the decay.

Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.

[3]

Markscheme

ALTERNATIVE 1

energy ratio expressed in terms of momentum, e.g. $\frac{E_{α}}{E_{lead}} = \frac{p_{α}^{2} / 2 m_{α}}{p_{lead}^{2} / 2 m_{lead}}$ ✓

$p_{α} = p_{lead}$ hence $\frac{E_{α}}{E_{lead}} = \frac{m_{lead}}{m_{α}}$ ✓

$\frac{m_{lead}}{m_{α}} ≃ \frac{206}{4} = 51.5 \Rightarrow E_{α} = 51.5 E_{lead}$ «so $α$ has a much greater KE»

OR

$m_{lead}$ «much» greater than $m_{alpha}$ «so $α$ has a much greater KE» ✓

ALTERNATIVE 2

alpha particle and lead particle have equal and opposite momenta ✓

so their velocities are inversely proportional to mass ✓

but KE ∝ v² «so $α$ has a much greater KE» ✓

Examiners report

Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.

(b.iii)

In the decay of polonium-210, alpha emission can be followed by the emission of a gamma photon.

State and explain whether the alpha particle or gamma photon will cause greater ionization in the surrounding material.

[2]

Markscheme

alpha particle ✓

is electrically charged hence more likely to interact with electrons «in the surrounding material» ✓

Examiners report

Many did not interpret the stem correctly and failed to compare the ionisation of alpha particles versus gamma rays.

a.

[Maximum mark: 2]

22N.2.SL.TZ0.6a

(a)

Outline, by reference to nuclear binding energy, why the mass of a nucleus is less than the sum of the masses of its constituent nucleons.

[2]

Markscheme

according to ΔE = Δmc² / identifies mass energy equivalence ✓

energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»

OR

energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓

Accept protons and neutrons.

Examiners report

Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.

b.i.

[Maximum mark: 2]

22N.2.SL.TZ0.6b.i

(b.i)

Calculate, in MeV, the energy released in this decay.

[2]

Markscheme

(m_polonium− m_lead − m_α)c² OR (209.93676 − 205.92945 − 4.00151)

OR

mass difference = 5.8 × 10⁻³ ✓

conversion to MeV using 931.5 to give 5.4 «MeV» ✓

Allow ECF from MP1.

Award [2] for a BCA.

Award [1] for 8.6 x 10⁻¹³ J.

Examiners report

Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c² as they were not sure of the significance of the unit of u.

b.ii.

[Maximum mark: 3]

22N.2.SL.TZ0.6b.ii

(b.ii)

The polonium nucleus was stationary before the decay.

Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.

[3]

Markscheme

ALTERNATIVE 1

energy ratio expressed in terms of momentum, e.g. $\frac{E_{α}}{E_{lead}} = \frac{p_{α}^{2} / 2 m_{α}}{p_{lead}^{2} / 2 m_{lead}}$ ✓

$p_{α} = p_{lead}$ hence $\frac{E_{α}}{E_{lead}} = \frac{m_{lead}}{m_{α}}$ ✓

$\frac{m_{lead}}{m_{α}} ≃ \frac{206}{4} = 51.5 \Rightarrow E_{α} = 51.5 E_{lead}$ «so $α$ has a much greater KE»

OR

$m_{lead}$ «much» greater than $m_{alpha}$ «so $α$ has a much greater KE» ✓

ALTERNATIVE 2

alpha particle and lead particle have equal and opposite momenta ✓

so their velocities are inversely proportional to mass ✓

but KE ∝ v² «so $α$ has a much greater KE» ✓

Examiners report

Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.

b.iii.

[Maximum mark: 2]

22N.2.SL.TZ0.6b.iii

(b.iii)

In the decay of polonium-210, alpha emission can be followed by the emission of a gamma photon.

State and explain whether the alpha particle or gamma photon will cause greater ionization in the surrounding material.

[2]

Markscheme

alpha particle ✓

is electrically charged hence more likely to interact with electrons «in the surrounding material» ✓

Examiners report

Many did not interpret the stem correctly and failed to compare the ionisation of alpha particles versus gamma rays.

a.

[Maximum mark: 1]

22N.2.SL.TZ0.a

State the initial acceleration of the raindrop.

[1]

Markscheme

g OR 9.81 «m s⁻²» OR acceleration of gravity/due to free fall ✓

Accept 10 «m s⁻²».

Ignore sign.

Do not accept bald “gravity”.

Accept answer that indicates tangent of the graph at time t=0.

Examiners report

A nice introductory question answered correctly by most candidates. Most answers quoted the data booklet value, with a few 10's or 9.8's, or the answer in words. Very few lost the mark by just stating gravity, or zero.

a.i.

[Maximum mark: 2]

22N.2.SL.TZ0.a.i

Draw, on the axes, a graph to show the variation with t of the displacement of particle Q.

[2]

Markscheme

oscillation in antiphase ✓

smaller amplitude than P ✓

Examiners report

Although there were good answers which scored full marks, there were a significant number of wrong answers where the amplitude was the same or not consistent throughout, or the wave drawn was not in antiphase of the original sketch.

a.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.a.ii

Calculate the speed of waves on the string.

[2]

Markscheme

wavelength $= \frac{2}{3} \times 0.80 = 0.53$ «m» ✓

speed $= \frac{0.53}{2.8 \times 10^{- 3}} = 190$ «m s⁻¹» ✓

Allow ECF from incorrect wavelength.

Examiners report

This was well answered, particularly MP1 to determine the wavelength, although several candidates misinterpreted the unit of time and obtained a very small value for the velocity of the wave.

b.

[Maximum mark: 3]

22N.2.SL.TZ0.b

Explain, by reference to the vertical forces, how the raindrop reaches a constant speed.

[3]

Markscheme

Identification of air resistance/drag force «acting upwards» ✓

«that» increases with speed ✓

«until» weight and air resistance cancel out

OR

net force/acceleration becomes zero ✓

A statement as “air resistance increases with speed” scores MP1 and MP2.

Examiners report

This was very well answered with most candidates scoring 3. The MP usually missed in candidates scoring 2 marks was MP2, to justify the variation of the magnitude of air resistance, although that rarely happened.

b.i.

[Maximum mark: 1]

22N.2.SL.TZ0.b.i

Calculate, in mA, the current in the resistor.

[1]

Markscheme

$\frac{1.47}{50.0} = 29.4$ «mA» ✓

Examiners report

Very well answered, with most candidates successfully answering in the required unit of mA.

b.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.b.ii

The water is heated. Explain why the quantity of air in the storage tank decreases.

[2]

Markscheme

ALTERNATIVE 1

«constant p and V imply» nT = const ✓

T increases hence n decreases ✓

ALTERNATIVE 2

«constant p and n imply» V is proportional to T / air expands as it is heated ✓

«original» air occupies a greater volume OR some air leaves through opening ✓

MP2 in ALT 2 must come from expansion of air, not from expansion of water.

Award [0] for an answer based on expansion of water.

Award [1] max for an answer based on convection currents.

Examiners report

It was surprising to see a large number of answers based on the expansion of water, as the stem of the question clearly states that the level of water remains constant. Most successful candidates scored by quoting pV constant so concluding with the inverse relationship of n and T, others also managed to score by explaining that the volume of air increases and therefore must go out through the opening. Answers based on convection currents were given partial credit.

b.iii.

[Maximum mark: 2]

22N.2.SL.TZ0.b.iii

In the decay of polonium-210, alpha emission can be followed by the emission of a gamma photon.

State and explain whether the alpha particle or gamma photon will cause greater ionization in the surrounding material.

[2]

Markscheme

alpha particle ✓

is electrically charged hence more likely to interact with electrons «in the surrounding material» ✓

Examiners report

Many did not interpret the stem correctly and failed to compare the ionisation of alpha particles versus gamma rays.

c.

[Maximum mark: 2]

22N.2.SL.TZ0.c

The standing wave on the string creates a travelling sound wave in the surrounding air.

Outline two differences between a standing wave and a travelling wave.

[2]

Markscheme

travelling waves transfer energy OR standing waves don’t ✓

amplitude of oscillation varies along a standing wave OR is constant along a travelling wave ✓

standing waves have nodes and antinodes OR travelling waves don’t ✓

points in an internodal region have same phase in standing waves OR different phase in travelling waves ✓

Examiners report

This question was answered well. Students showed to be familiar with the differences between standing and travelling waves. In SL they had to identify two differences, so that proved to be more challenging.

c.i.

[Maximum mark: 3]

22N.2.SL.TZ0.c.i

Determine the energy transferred to the air during the first 3.0 s of motion. State your answer to an appropriate number of significant figures.

[3]

Markscheme

«loss in» GPE = 3.4 × 10⁻⁵ × 9.81 × 21 «= 7.0 × 10⁻³» «J»

OR

«gain in» KE = 0.5 × 3.4 × 10⁻⁵ × 9.0² «= 1.4 × 10⁻³» «J» ✓

energy transferred to air «=7.0 × 10⁻³ − 1.4 × 10⁻³» = 5.6 × 10⁻³» «J» ✓

any calculated answer to 2 sf ✓

Allow [1] through the use of kinematics assuming constant acceleration.

Allow ECF from MP1.

Examiners report

Generally well answered, although several candidates lost a mark, usually as POT (power of ten) by quoting the value in kg leading to an answer of 5.3 J. Most candidates were able to score MP3 by rounding their calculation to two significant figures.

c.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.c.ii

The speed of sound in air is 340 m s⁻¹ and in water it is 1500 m s⁻¹.

Discuss whether the sound wave can enter the water.

[2]

Markscheme

ALTERNATIVE 1

critical angle $= 13 °$ «from $\sin θ_{c} = \frac{340}{1500}$ » ✓

the angle of incidence is greater than $θ_{c}$ hence the sound can’t enter water ✓

ALTERNATIVE 2

$\sin θ_{r} = \frac{1500}{340} \sin 30 °$ ✓

sine value greater than one hence the sound can’t enter water ✓

Conclusion must be justified, award [0] for BCA.

Examiners report

Surprisingly well answered as it was sound from air to water, rather than light from air to glass. A mixture of approaches but probably the most common was to calculate a sine value of over 1. Some went about calculating the critical angle but nowhere near as many.

3.

[Maximum mark: 1]

23M.1A.SL.TZ1.3

A car travels clockwise around a circular track of radius R. What is the magnitude of displacement from X to Y?

A. $R \frac{3 π}{2}$

B. $R \frac{π}{2}$

C. $R \sqrt{2}$

D. $R$

[1]

Markscheme

C

30.

[Maximum mark: 1]

23M.1A.SL.TZ1.30

Light of intensity 500 W m⁻² is incident on concrete and on snow. 300 W m⁻² is reflected from the
concrete and 400 W m⁻² is reflected from the snow.

What is $\frac{albedo of concrete}{albedo of snow}$ ?

A. $\frac{1}{2}$

B. $\frac{3}{4}$

C. $\frac{4}{3}$

D. 2

[1]

Markscheme

B

4.

[Maximum mark: 1]

23M.1A.SL.TZ1.4

A car accelerates uniformly. The car passes point X at time t₁ with velocity v₁ and point Y at time t₂ with velocity v₂. The distance XY is s.

The following expressions are proposed for the magnitude of its acceleration a:

I. $a = \frac{2 s}{{(t_{2} - t_{1})}^{2}}$

II. $a = \frac{{v_{2}}^{2} - {v_{1}}^{2}}{2 s}$

III. $a = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

Which is correct?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

[1]

Markscheme

C

5.

[Maximum mark: 1]

23M.1A.SL.TZ1.5

A ball is projected at an angle to the horizonal on Earth reaching a maximum height H and a maximum range R. The same ball is projected at the same angle and speed on a planet where the acceleration due to gravity is three times that on Earth. Resistance effects are negligible.

What is the maximum range and the maximum height reached on that planet?

	Maximum range	Maximum height reached
A.	$\frac{R}{3}$	$\frac{H}{3}$
B.	$R$	$\frac{H}{3}$
C.	$R$	$\frac{H}{9}$
D.	$\frac{R}{3}$	$\frac{H}{9}$

[1]

Markscheme

A

1.

[Maximum mark: 1]

23M.1A.SL.TZ2.1

A rocket travels a distance of 3 km in 10 s.

What is the order of magnitude of $\frac{the speed of the rocket}{the speed of light in a vacuum}$ ?

A. −5

B. −6

C. −7

D. −8

[1]

Markscheme

B

10.

[Maximum mark: 1]

23M.1A.SL.TZ2.10

A balloon of volume V contains 10 mg of an ideal gas at a pressure P. An additional mass of the gas is added without changing the temperature of the balloon. This change causes the volume to increase to 2V and the pressure to increase to 3P.

What is the mass of gas added to the balloon?

A. 5 mg

B. 15 mg

C. 50 mg

D. 60 mg

[1]

Markscheme

C

11.

[Maximum mark: 1]

23M.1A.SL.TZ2.11

A pipe containing air is closed at one end and open at the other. The third harmonic standing wave for this pipe has a frequency of 150 Hz.

What other frequency is possible for a standing wave in this pipe?

A. 25 Hz

B. 50 Hz

C. 75 Hz

D. 300 Hz

[1]

Markscheme

B

12.

[Maximum mark: 1]

23M.1A.SL.TZ2.12

A longitudinal wave is travelling through a medium. The variation with distance d of the displacement $x$ of the particles in the medium at time t is shown.

Which point is at the centre of a compression?

[1]

Markscheme

A

13.

[Maximum mark: 1]

23M.1A.SL.TZ2.13

A mass is oscillating with simple harmonic motion. At time t, the acceleration is at a positive maximum.

What are the displacement and velocity of the mass at time t?

	Displacement	Velocity
A.	positive maximum	zero
B.	negative maximum	zero
C.	positive maximum	negative maximum
D.	negative maximum	negative maximum

[1]

Markscheme

B

14.

[Maximum mark: 1]

23M.1A.SL.TZ2.14

An electromagnetic wave enters a medium of lower refractive index.

Three statements are made:

I. The wavelength of the wave has increased.
II. The frequency of the wave has decreased.
III. The speed of the wave has increased.

What is true about the properties of the wave?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

[1]

Markscheme

B

15.

[Maximum mark: 1]

23M.1A.SL.TZ2.15

Three point charges, +Q, +Q and −Q, are fixed at the three corners of a square.

What is the direction of the electric field at the fourth corner?

[1]

Markscheme

B

16.

[Maximum mark: 1]

23M.1A.SL.TZ2.16

A longitudinal wave is travelling through a medium. The variation with distance d of the displacement $x$ of the particles in the medium at time t is shown.

Which point is at the centre of a compression?

[1]

Markscheme

A

17.

[Maximum mark: 1]

23M.1A.SL.TZ2.17

A variable resistor is connected to a cell with emf ε and internal resistance r as shown. When the current in the circuit is I, the potential difference measured across the terminals of the cell is V.

The resistance of the variable resistor is doubled.

What is true about the current and the potential difference?

	Current	Potential difference
A.	greater than $\frac{I}{2}$	greater than V
B.	less than $\frac{I}{2}$	greater than V
C.	greater than $\frac{I}{2}$	equal to V
D.	less than $\frac{I}{2}$	equal to V

[1]

Markscheme

A

18.

[Maximum mark: 1]

23M.1A.SL.TZ2.18

A negatively charged sphere is falling through a magnetic field.

What is the direction of the magnetic force acting on the sphere?

A. To the left of the page

B. To the right of the page

C. Out of the page

D. Into the page

[1]

Markscheme

D

19.

[Maximum mark: 1]

23M.1A.SL.TZ2.19

Three point charges, +Q, +Q and −Q, are fixed at the three corners of a square.

What is the direction of the electric field at the fourth corner?

[1]

Markscheme

B

2.

[Maximum mark: 1]

23M.1A.SL.TZ2.2

The variation with time of the displacement of an object is shown.

What are the average speed and average velocity of the object over the 10 s time interval?

	Average speed / m s⁻¹	Average velocity / m s⁻¹
A.	0.8	0.8
B.	0.8	1.2
C.	1.2	0.8
D.	1.2	1.2

[1]

Markscheme

C

20.

[Maximum mark: 1]

23M.1A.SL.TZ2.20

Two resistors of equal resistance R are connected with two cells of emf ε and 2ε. Both cells have negligible internal resistance.

What is the current in the resistor labelled X?

A. $\frac{ε}{2 R}$

B. $\frac{3 ε}{2 R}$

C. $\frac{ε}{R}$

D. $\frac{3 ε}{R}$

[1]

Markscheme

C

21.

[Maximum mark: 1]

23M.1A.SL.TZ2.21

A negatively charged sphere is falling through a magnetic field.

What is the direction of the magnetic force acting on the sphere?

A. To the left of the page

B. To the right of the page

C. Out of the page

D. Into the page

[1]

Markscheme

D

22.

[Maximum mark: 1]

23M.1A.SL.TZ2.22

A student measures the count rate of a radioactive sample with time in a laboratory. The background count in the laboratory is 30 counts per second.

Count rate / counts per second	Time / s
150	0
90	20

What is the time at which the student measures a count rate of 45 counts per second?

A. 30 s

B. 40 s

C. 60 s

D. 80 s

[1]

Markscheme

C

23.

[Maximum mark: 1]

23M.1A.SL.TZ2.23

Three statements about the binding energy are provided.

I. The binding energy is the energy required to completely separate the nucleons.
II. The binding energy is equivalent, in units of energy, to the mass defect when a nucleus is formed from its nucleons.
III. The binding energy is the energy released when a nucleus is formed from its nucleons.

Which statements are true?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

[1]

Markscheme

D

24.

[Maximum mark: 1]

23M.1A.SL.TZ2.24

A nucleus of platinum (Pt) undergoes alpha decay to form an osmium (Os) nucleus as represented by the following reaction.

${}_{78}^{175}{Pt}$ → Os + alpha particle

What are the number of protons and the number of neutrons in the osmium nucleus?

	Number of protons	Number of neutrons
A.	74	93
B.	76	93
C.	74	95
D.	76	95

[1]

Markscheme

D

25.

[Maximum mark: 1]

23M.1A.SL.TZ2.25

A car engine has a useful power output of 20 kW and an efficiency of 50 %. The engine consumes 1 × 10⁻⁵ m³ of fuel every second. What is the energy density of the fuel?

A. 2 MJ m⁻³

B. 4 MJ m⁻³

C. 2 GJ m⁻³

D. 4 GJ m⁻³

[1]

Markscheme

D

26.

[Maximum mark: 1]

23M.1A.SL.TZ2.26

The gravitational field strength at the surface of the Earth is often taken to be 9.8 N kg⁻¹.

The use of this value to calculate the weight of an object above the surface of the Earth is

A. a paradigm shift in our understanding of gravity.

B. an attempt to model gravitational fields.

C. an outcome from a peer review.

D. an approximation used for estimation purposes.

[1]

Markscheme

D

27.

[Maximum mark: 1]

23M.1A.SL.TZ2.27

A planet has an albedo of 0.30. A simplified energy balance for the planet is shown.

What is the intensity radiated by the surface of the planet?

A. 70 W m⁻²

B. 90 W m⁻²

C. 100 W m⁻²

D. 130 W m⁻²

[1]

Markscheme

D

3.

[Maximum mark: 1]

23M.1A.SL.TZ2.3

A stone is thrown vertically up from the top of a cliff with a velocity v at time t = 0. Air resistance is negligible.

What is the variation with time of the velocity of the stone until it hits the ground?

[1]

Markscheme

A

30.

[Maximum mark: 1]

23M.1A.SL.TZ2.30

A planet has an albedo of 0.30. A simplified energy balance for the planet is shown.

What is the intensity radiated by the surface of the planet?

A. 70 W m⁻²

B. 90 W m⁻²

C. 100 W m⁻²

D. 130 W m⁻²

[1]

Markscheme

D

4.

[Maximum mark: 1]

23M.1A.SL.TZ2.4

Ball 1 is released at rest from the top of a building. At the same instant in time, Ball 2 is projected horizontally from the same height. The effect of air resistance is negligible.

Which statement is true?

A. The velocity on impact with the ground is the same for both balls.

B. The time taken to hit the ground is greater for Ball 2.

C. The speed on impact with the ground is the same for both balls.

D. The velocity on impact with the ground is greater for Ball 2.

[1]

Markscheme

D

1.

[Maximum mark: 15]

23M.2.SL.TZ2.1

A toy rocket is made from a plastic bottle that contains some water.

Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.

The air–water mixture is called the propellant.

The variation with time of the vertical velocity of the bottle is shown.

The bottle reaches its highest point at time T₁ on the graph and returns to the ground at time T₂. The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.

(a)

Estimate, using the graph, the maximum height of the bottle.

[3]

Markscheme

ALTERNATIVE 1

Attempt to count squares ✓

Area of one square found ✓

7.2 «m» (accept 6.4 – 7.4 m) ✓

ALTERNATIVE 2

Uses area equation for either triangle ✓

Correct read offs for estimate of area of triangle ✓

7.2 «m» (accept 6.4 – 7.4) ✓

(b)

Estimate the acceleration of the bottle when it is at its maximum height.

[2]

Markscheme

Attempt to calculate gradient of line at t = 1.2 s ✓

«−» 9.8 «m s⁻²» (accept 9.6 − 10.0) ✓

(c)

The bottle bounces when it returns to the ground.

(c.i)

Calculate the fraction of the kinetic energy of the bottle that remains after the bounce.

[2]

Markscheme

Attempt to evaluate KE ratio as ${(\frac{V_{final}}{V_{initial}})}^{2}$ ✓

« ${(\frac{4.5}{10})}^{2}$ =» 0.20 OR 20 % OR $\frac{1}{5}$ ✓

Accept ± 0.5 velocity values from graph

(c.ii)

The mass of the bottle is 27 g and it is in contact with the ground for 85 ms.

Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.

[3]

Markscheme

Attempt to use force = momentum change ÷ time ✓

«= $\frac{(4.5 + 10) \times 0.027}{85 \times 10^{- 3}}$ = 4.6»

Force = «4.6 + 0.3» 4.9 «N» ✓

Any answer to 2sf ✓

Accept ± 0.5 velocity values from graph

(d)

After a second bounce, the bottle rotates about its centre of mass. The bottle rotates at 0.35 revolutions per second.

The centre of mass of the bottle is halfway between the base and the top of the bottle. Assume that the velocity of the centre of mass is zero.

Calculate the linear speed of the top of the bottle.

[3]

Markscheme

ALTERNATIVE 1

ω = 2 $π$ (0.35) «=2.20 rad s⁻¹» ✓
Use of v = 0.14ω ✓
0.31 «m s⁻¹» ✓

ALTERNATIVE 2

T = $\frac{1}{0.35}$ «= 2.9 s» ✓

V = $\frac{2 π (0.14)}{T}$ OR $\frac{v^{2}}{0.14}$ = $\frac{4 π^{2} (0.14)}{T^{2}}$ ✓

v = 0.31 «m s⁻¹» ✓

Award [3] for BCA

(e)

The maximum height reached by the bottle is greater with an air–water mixture than with only high-pressure air in the bottle.

Assume that the speed at which the propellant leaves the bottle is the same in both cases.

Explain why the bottle reaches a greater maximum height with an air–water mixture.

[2]

Markscheme

Mass «leaving the bottle per second» will be larger for air–water ✓

the momentum change/force is greater ✓

Allow opposite argument for air only

i.

[Maximum mark: 2]

23M.2.SL.TZ2.11bi

(i)

Determine which star will appear to move more.

[2]

Markscheme

Star Y ✓

because parallax angle is greater OR star Y is closer «and that means movement relative to distant stars is greater» ✓

Allow reverse argument for star X

ii.

[Maximum mark: 1]

23M.2.SL.TZ2.11bii

(ii)

Calculate, in m, the distance to star X.

[1]

Markscheme

«distance = $(\frac{1}{0.019})$ × 3.26 × 9.46 × 10¹⁵»

1.6 × 10¹⁸ «m» ✓

iii.

[Maximum mark: 2]

23M.2.SL.TZ2.11biii

(iii)

Determine the ratio $\frac{luminosity of star X}{luminosity of star Y}$ .

[2]

Markscheme

$\frac{Luminosity of Star X}{Luminosity of Star Y}$ = $\frac{b_{x} d_{x}^{2}}{b_{y} d_{y}^{2}}$ ✓

= 10.8 ≈ 11 ✓

Award MP1 if ratio shown with distance or parallax angle.

Award MP1 for any correct substitution into ratio expression

Award [2] for BCA

Allow ECF for incorrect distances from b(i) or b(ii).

a.

[Maximum mark: 1]

23M.2.SL.TZ2.12a

(a)

State the main element that is undergoing nuclear fusion in star C.

[1]

Markscheme

Hydrogen ✓

b.

[Maximum mark: 2]

23M.2.SL.TZ2.12b

(b)

Explain why star B has a greater surface area than star A.

[2]

Markscheme

stars have same/similar L AND star B has lower T ✓

correct reference to luminosity formula (Lα AT⁴) ✓

MP1 Allow reverse argument i.e., star A has higher T

c.

[Maximum mark: 2]

23M.2.SL.TZ2.12c

(c)

White dwarfs with similar volumes to each other are shown on the HR diagram.

Sketch, on the HR diagram, to show the possible positions of other white dwarf stars with similar volumes to those marked on the HR diagram.

[2]

Markscheme

Any evidence of correct identification that three dots bottom left represent white dwarfs ✓

line passing through all 3 white dwarfs OR line continuing from 3 white dwarfs with approximately same gradient, in either direction ✓

Award MP2 if no line drawn through the three dots but just beyond them in either direction

a.

[Maximum mark: 1]

23M.2.SL.TZ2.17a

(a)

State the main element that is undergoing nuclear fusion in star C.

[1]

Markscheme

Hydrogen ✓

b.

[Maximum mark: 2]

23M.2.SL.TZ2.17b

(b)

Explain why star B has a greater surface area than star A.

[2]

Markscheme

stars have same/similar L AND star B has lower T ✓

correct reference to luminosity formula (Lα AT⁴) ✓

MP1 Allow reverse argument i.e., star A has higher T

c.

[Maximum mark: 2]

23M.2.SL.TZ2.17c

(c)

White dwarfs with similar volumes to each other are shown on the HR diagram.

Sketch, on the HR diagram, to show the possible positions of other white dwarf stars with similar volumes to those marked on the HR diagram.

[2]

Markscheme

Any evidence of correct identification that three dots bottom left represent white dwarfs ✓

line passing through all 3 white dwarfs OR line continuing from 3 white dwarfs with approximately same gradient, in either direction ✓

Award MP2 if no line drawn through the three dots but just beyond them in either direction

a.

[Maximum mark: 3]

23M.2.SL.TZ2.1a

(a)

Estimate, using the graph, the maximum height of the bottle.

[3]

Markscheme

ALTERNATIVE 1

Attempt to count squares ✓

Area of one square found ✓

7.2 «m» (accept 6.4 – 7.4 m) ✓

ALTERNATIVE 2

Uses area equation for either triangle ✓

Correct read offs for estimate of area of triangle ✓

7.2 «m» (accept 6.4 – 7.4) ✓

b.

[Maximum mark: 2]

23M.2.SL.TZ2.1b

(b)

Estimate the acceleration of the bottle when it is at its maximum height.

[2]

Markscheme

Attempt to calculate gradient of line at t = 1.2 s ✓

«−» 9.8 «m s⁻²» (accept 9.6 − 10.0) ✓

i.

[Maximum mark: 1]

23M.2.SL.TZ2.1ci

(i)

State the unit for k.

[1]

Markscheme

N²m² OR kg²m⁴s⁻⁴ OR N³Pa⁻¹ ✓

Award [0] if they convert to base units incorrectly.

ii.

[Maximum mark: 3]

23M.2.SL.TZ2.1cii

(ii)

The mass of the bottle is 27 g and it is in contact with the ground for 85 ms.

Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.

[3]

Markscheme

Attempt to use force = momentum change ÷ time ✓

«= $\frac{(4.5 + 10) \times 0.027}{85 \times 10^{- 3}}$ = 4.6»

Force = «4.6 + 0.3» 4.9 «N» ✓

Any answer to 2sf ✓

Accept ± 0.5 velocity values from graph

d.

[Maximum mark: 3]

23M.2.SL.TZ2.1d

(d)

After a second bounce, the bottle rotates about its centre of mass. The bottle rotates at 0.35 revolutions per second.

The centre of mass of the bottle is halfway between the base and the top of the bottle. Assume that the velocity of the centre of mass is zero.

Calculate the linear speed of the top of the bottle.

[3]

Markscheme

ALTERNATIVE 1

ω = 2 $π$ (0.35) «=2.20 rad s⁻¹» ✓
Use of v = 0.14ω ✓
0.31 «m s⁻¹» ✓

ALTERNATIVE 2

T = $\frac{1}{0.35}$ «= 2.9 s» ✓

V = $\frac{2 π (0.14)}{T}$ OR $\frac{v^{2}}{0.14}$ = $\frac{4 π^{2} (0.14)}{T^{2}}$ ✓

v = 0.31 «m s⁻¹» ✓

Award [3] for BCA

i.

[Maximum mark: 3]

23M.2.SL.TZ2.1di

(i)

Calculate the absolute uncertainty in $F_{\max}^{3}$ for Δp = 30 kPa. State an appropriate number of significant figures for your answer.

[3]

Markscheme

15 % seen anywhere ✓

«Δ(F³) =» 39.4 × 10⁵ × 0.15 = 5.9 × 10⁵ ✓

±6 × 10⁵ ✓

MP1 is for the propagation of 5 %. It can be shown differently, e.g. 3 × 5% Allow students to use 40 × 10⁵ (from the graph).

Award MP3 for any uncertainty rounded to 1 significant digit

Award [3] for a BCA.

Allow ECF from MP1 and MP2

ii.

[Maximum mark: 1]

23M.2.SL.TZ2.1dii

(ii)

Draw the absolute uncertainty determined in part (d)(i) as an error bar on the graph.

[1]

Markscheme

error bar drawn at 30 kPa from 34 × 10⁵ to 46 × 10⁵ N³ ✓

Allow ± half square on each side of the bar or one square overall (± 2 × 10⁵)

Allow ECF from d(i).

iii.

[Maximum mark: 1]

23M.2.SL.TZ2.1diii

(iii)

Explain why the new hypothesis is supported.

[1]

Markscheme

a «straight» line can be drawn that passes through origin ✓

e.

[Maximum mark: 2]

23M.2.SL.TZ2.1e

(e)

The maximum height reached by the bottle is greater with an air–water mixture than with only high-pressure air in the bottle.

Assume that the speed at which the propellant leaves the bottle is the same in both cases.

Explain why the bottle reaches a greater maximum height with an air–water mixture.

[2]

Markscheme

Mass «leaving the bottle per second» will be larger for air–water ✓

the momentum change/force is greater ✓

Allow opposite argument for air only

2.

[Maximum mark: 7]

23M.2.SL.TZ2.2

Pressure p, volume V and temperature T are measured for a fixed mass of gas.
T is measured in degrees Celsius.

The graph shows the variation of pV with T.

The mass of a molecule of the gas is 4.7 × 10⁻²⁶ kg.

(a)

State the unit for pV in fundamental SI units.

[1]

Markscheme

kg m² s⁻² ✓

(b)

Determine, using the graph, whether the gas acts as an ideal gas.

[3]

Markscheme

ALTERNATIVE 1

Graph shown is a straight line/linear
OR
expected graph should be a straight line/linear ✓

If ideal then T intercept must be at T = −273 °C ✓

Use of y = mx+c to show that x = −273 °C when y = 0 ✓
(hence ideal)

ALTERNATIVE 2

Calculates $\frac{p V}{T}$ for two different points ✓
Obtains 1.50 «J K⁻¹» for both ✓

States that for ideal gas $\frac{p V}{T} = n R$ which is constant and concludes that gas is ideal ✓

(c)

Calculate, in g, the mass of the gas.

[3]

Markscheme

Use of $n = \frac{p V}{R T}$ OR $N = \frac{p V}{k T}$ ✓

Mass of gas = n × N_A × mass of molecule
OR
Mass of gas = N × mass of molecule ✓

5.1 «g» ✓

a.

[Maximum mark: 2]

23M.2.SL.TZ2.20a

(a)

Determine the critical density of the universe using a Hubble constant value of 73 km s⁻¹ Mpc⁻¹.

[2]

Markscheme

H «= $\frac{73 \times 10^{3}}{3.26 \times 10^{6} \times 9.46 \times 10^{15}}$ » = 2.37 × 10⁻¹⁸ «s⁻¹» ✓

ρ_c = « $\frac{3 \times {(2.37 \times 10^{- 18})}^{2}}{8 π \times 6.67 \times 10^{- 11}}$ =» 1.0 × 10⁻²⁶ «kg m⁻³» ✓

Award [2] for BCA

Allow ECF from MP1

Award [1] for 9.5 × 10¹⁸ or 9.5 × 10¹²

a.

[Maximum mark: 1]

23M.2.SL.TZ2.2a

(a)

State one other measurement that the student will need to make.

[1]

Markscheme

final temperature of equilibrium/water/cube OR mass of water/cube ✓

Do not award mark if any additional incorrect measurement is included

Accept temperature change if water or cube specified

b.

[Maximum mark: 3]

23M.2.SL.TZ2.2b

(b)

Determine, using the graph, whether the gas acts as an ideal gas.

[3]

Markscheme

ALTERNATIVE 1

Graph shown is a straight line/linear
OR
expected graph should be a straight line/linear ✓

If ideal then T intercept must be at T = −273 °C ✓

Use of y = mx+c to show that x = −273 °C when y = 0 ✓
(hence ideal)

ALTERNATIVE 2

Calculates $\frac{p V}{T}$ for two different points ✓
Obtains 1.50 «J K⁻¹» for both ✓

States that for ideal gas $\frac{p V}{T} = n R$ which is constant and concludes that gas is ideal ✓

c.

[Maximum mark: 3]

23M.2.SL.TZ2.2c

(c)

Calculate, in g, the mass of the gas.

[3]

Markscheme

Use of $n = \frac{p V}{R T}$ OR $N = \frac{p V}{k T}$ ✓

Mass of gas = n × N_A × mass of molecule
OR
Mass of gas = N × mass of molecule ✓

5.1 «g» ✓

3.

[Maximum mark: 9]

23M.2.SL.TZ2.3

Blue light of wavelength $λ$ is incident on a double slit. Light from the double slit falls on a screen. A student measures the distance between nine successive fringes on the screen to be 15 cm.

The separation of the double slit is 60 µm; the double slit is 2.5 m from the screen.

(a)

Explain the pattern seen on the screen.

[3]

Markscheme

Mention of interference / superposition ✓

Bright fringe occurs when light from the slits arrives in phase ✓

Dark fringe occurs when light from the slits arrives 180°/ $π$ out of phase ✓

(b)

Calculate, in nm, $λ$ .

[3]

Markscheme

s = $\frac{0.15}{8}$ OR = 0.0188 «m» ✓

use of $λ = \frac{d s}{D}$ ✓

450 «nm» ✓

(c)

The student changes the light source to one that emits two colours:
• blue light of wavelength $λ$ , and
• red light of wavelength 1.5 $λ$ .

Predict the pattern that the student will see on the screen.

[3]

Markscheme

Blue fringe is unchanged ✓

Red fringes are farther apart than blue ✓

By a factor of 1.5 ✓

At some point/s the fringes coincide/are purple ✓

a.

[Maximum mark: 3]

23M.2.SL.TZ2.3a

(a)

Explain the pattern seen on the screen.

[3]

Markscheme

Mention of interference / superposition ✓

Bright fringe occurs when light from the slits arrives in phase ✓

Dark fringe occurs when light from the slits arrives 180°/ $π$ out of phase ✓

b.

[Maximum mark: 3]

23M.2.SL.TZ2.3b

(b)

Calculate, in nm, $λ$ .

[3]

Markscheme

s = $\frac{0.15}{8}$ OR = 0.0188 «m» ✓

use of $λ = \frac{d s}{D}$ ✓

450 «nm» ✓

i.

[Maximum mark: 3]

23M.2.SL.TZ2.3bi

(i)

Calculate, in nm, $λ$ .

[3]

Markscheme

s = $\frac{0.15}{8}$ OR = 0.0188 «m» ✓

use of $λ = \frac{d s}{D}$ ✓

450 «nm» ✓

ii.

[Maximum mark: 2]

23M.2.SL.TZ2.3bii

(ii)

The student moves the screen closer to the double slit and repeats the measurements. The instruments used to make the measurements are unchanged.

Discuss the effect this movement has on the fractional uncertainty in the value of $λ$ .

[2]

Markscheme

«As the measurements decrease» the fractional uncertainty in D/s increases. ✓

«Fractional uncertainties are additive here» so fractional uncertainty in $λ$ increases ✓

Answers can be described in symbols e.g. Δs/s

c.

[Maximum mark: 3]

23M.2.SL.TZ2.3c

(c)

The student changes the light source to one that emits two colours:
• blue light of wavelength $λ$ , and
• red light of wavelength 1.5 $λ$ .

Predict the pattern that the student will see on the screen.

[3]

Markscheme

Blue fringe is unchanged ✓

Red fringes are farther apart than blue ✓

By a factor of 1.5 ✓

At some point/s the fringes coincide/are purple ✓

4.

[Maximum mark: 12]

23M.2.SL.TZ2.4

An electrically heated pad is designed to keep a pet warm.

The pad is heated using a resistor that is placed inside the pad. The dimensions of the resistor are shown on the diagram. The resistor has a resistance of 4.2 Ω and a total length of 1.25 m.

diagram not to scale

When there is a current in the resistor, the temperature in the pad changes from a room temperature of 20 °C to its operating temperature at 35 °C.

(a)

The designers state that the energy transferred by the resistor every second is 15 J.

Calculate the current in the resistor.

[1]

Markscheme

I = « $\sqrt{\frac{P}{R}}$ =» 1.9 «A» ✓

(b)

The designers wish to make the resistor from carbon fibre.

The graph shows the variation with temperature, in Kelvin, of the resistivity of carbon fibre.

(b.i)

The resistor has a cross-sectional area of 9.6 × 10⁻⁶ m².

Show that a resistor made from carbon fibre will be suitable for the pad.

[3]

Markscheme

ALTERNATIVE 1 (Calculation of length)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $I = \frac{R A}{ρ}$ ✓
l = 1.3 − 1.4 «m» ✓

ALTERNATIVE 2 (Calculation of area)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $A = ρ \frac{I}{R}$ ✓

A = 8.3 − 9.5 × 10⁻⁶ «m²» ✓

ALTERNATIVE 3 (Calculation of resistance)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $R = ρ \frac{I}{A}$ ✓

R = 3.6 − 4.2 «Ω» ✓

ALTERNATIVE 4 (Calculation of resistivity)
Use of $ρ = \frac{R A}{I}$ ✓

$ρ$ = 3.2 × 10⁻⁵ «Ω m» ✓

Read off from graph 260 – 280 K ✓

(b.ii)

The power supply to the pad has a negligible internal resistance.

State and explain the variation in current in the resistor as the temperature of the pad increases.

[2]

Markscheme

«Resistivity and hence» resistance will decrease ✓

«Pd across pad will not change because internal resistance is negligible»
Current will increase ✓

(c)

When there is a current in the resistor, magnetic forces act between the resistor strips.

For the part of the resistor labelled RS,

(c.i)

outline the magnetic force acting on it due to the current in PQ.

[1]

Markscheme

«The force is» away from PQ/repulsive/to the right ✓

(c.ii)

state and explain the net magnetic force acting on it due to the currents in PQ and TU.

[2]

Markscheme

The magnetic fields «due to currents in PQ and TU» are in opposite directions
OR
There are two «repulsive» forces in opposite directions ✓

Net force is zero ✓

(d)

The design of the pad encloses the resistor in a material that traps air. The design also places the resistor close to the top surface of the pad.

Explain, with reference to thermal energy transfer, why the pad is designed in this way.

[3]

Markscheme

Air is a poor «thermal» conductor ✓

Lack of convection due to air not being able to move in material ✓

Appropriate statement about energy transfer between the pet, the resistor and surroundings ✓

The rate of thermal energy transfer to the top surface is greater than the bottom «due to thinner material» ✓

Accept air is a good insulator

a.

[Maximum mark: 1]

23M.2.SL.TZ2.4a

(a)

The designers state that the energy transferred by the resistor every second is 15 J.

Calculate the current in the resistor.

[1]

Markscheme

I = « $\sqrt{\frac{P}{R}}$ =» 1.9 «A» ✓

i.

[Maximum mark: 3]

23M.2.SL.TZ2.4bi

(i)

The resistor has a cross-sectional area of 9.6 × 10⁻⁶ m².

Show that a resistor made from carbon fibre will be suitable for the pad.

[3]

Markscheme

ALTERNATIVE 1 (Calculation of length)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $I = \frac{R A}{ρ}$ ✓
l = 1.3 − 1.4 «m» ✓

ALTERNATIVE 2 (Calculation of area)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $A = ρ \frac{I}{R}$ ✓

A = 8.3 − 9.5 × 10⁻⁶ «m²» ✓

ALTERNATIVE 3 (Calculation of resistance)
Read off from graph [2.8 − 3.2 × 10⁻⁵ Ω m]✓
Use of $R = ρ \frac{I}{A}$ ✓

R = 3.6 − 4.2 «Ω» ✓

ALTERNATIVE 4 (Calculation of resistivity)
Use of $ρ = \frac{R A}{I}$ ✓

$ρ$ = 3.2 × 10⁻⁵ «Ω m» ✓

Read off from graph 260 – 280 K ✓

ii.

[Maximum mark: 2]

23M.2.SL.TZ2.4bii

(ii)

The power supply to the pad has a negligible internal resistance.

State and explain the variation in current in the resistor as the temperature of the pad increases.

[2]

Markscheme

«Resistivity and hence» resistance will decrease ✓

«Pd across pad will not change because internal resistance is negligible»
Current will increase ✓

i.

[Maximum mark: 1]

23M.2.SL.TZ2.4ci

(i)

outline the magnetic force acting on it due to the current in PQ.

[1]

Markscheme

«The force is» away from PQ/repulsive/to the right ✓

ii.

[Maximum mark: 2]

23M.2.SL.TZ2.4cii

(ii)

state and explain the net magnetic force acting on it due to the currents in PQ and TU.

[2]

Markscheme

The magnetic fields «due to currents in PQ and TU» are in opposite directions
OR
There are two «repulsive» forces in opposite directions ✓

Net force is zero ✓

d.

[Maximum mark: 3]

23M.2.SL.TZ2.4d

(d)

The design of the pad encloses the resistor in a material that traps air. The design also places the resistor close to the top surface of the pad.

Explain, with reference to thermal energy transfer, why the pad is designed in this way.

[3]

Markscheme

Air is a poor «thermal» conductor ✓

Lack of convection due to air not being able to move in material ✓

Appropriate statement about energy transfer between the pet, the resistor and surroundings ✓

The rate of thermal energy transfer to the top surface is greater than the bottom «due to thinner material» ✓

Accept air is a good insulator

a.

[Maximum mark: 1]

23M.2.SL.TZ2.5a

(a)

Outline what is meant by an isotope.

[1]

Markscheme

«An atom with» the same number of protons AND different numbers of neutrons
OR
Same chemical properties AND different physical properties ✓

Do not allow just atomic number and mass number

i.

[Maximum mark: 1]

23M.2.SL.TZ2.5bi

(i)

mass number.

[1]

Markscheme

3 ✓

ii.

[Maximum mark: 1]

23M.2.SL.TZ2.5bii

(ii)

proton number.

[1]

Markscheme

2 ✓

e.

[Maximum mark: 2]

23M.2.SL.TZ2.5e

(e)

A beta-minus particle and an alpha particle have the same initial kinetic energy.

Outline why the beta-minus particle can travel further in air than the alpha particle.

[2]

Markscheme

Alphas have double charge «and so are better ionisers »✓

alphas have more mass and therefore slower «for same energy» ✓

so longer time/more likely to interact with the «atomic» electrons/atoms «and therefore better ionisers» ✓

Accept reverse argument in terms of betas travelling faster.

a.

[Maximum mark: 2]

23M.2.SL.TZ2.6a

(a)

Determine $\frac{g_{M}}{g_{P}}$ .

[2]

Markscheme

Work using g ∝ $\frac{m}{r^{2}}$ ✓

$\frac{g_{M}}{g_{P}} = \frac{m_{M}}{m_{P}} {(\frac{r_{P}}{r_{M}})}^{2}$ = 0.75 ✓

e.

[Maximum mark: 1]

23M.2.SL.TZ2.6e

(e)

Outline one reason why this model of a dancer is unrealistic.

[1]

Markscheme

one example specified eg friction, air resistance, mass distribution not modelled ✓

Award [1] for any reasonable physical parameter that is not consistent with the model

a.

[Maximum mark: 1]

23M.2.SL.TZ2.9a

(a)

Show, using the data, that the energy released in the decay of one magnesium-27 nucleus is about 2.62 MeV.

Mass of aluminium-27 atom = 26.98153 u
Mass of magnesium-27 atom = 26.98434 u
The unified atomic mass unit is 931.5 MeV c⁻².

[1]

Markscheme

(26.98434 - 26.98153) × 931.5

OR

2.6175 «MeV» seen ✓

a.

[Maximum mark: 3]

23M.2.SL.TZ2.a

Estimate, using the graph, the maximum height of the bottle.

[3]

Markscheme

ALTERNATIVE 1

Attempt to count squares ✓

Area of one square found ✓

7.2 «m» (accept 6.4 – 7.4 m) ✓

ALTERNATIVE 2

Uses area equation for either triangle ✓

Correct read offs for estimate of area of triangle ✓

7.2 «m» (accept 6.4 – 7.4) ✓

b.

[Maximum mark: 2]

23M.2.SL.TZ2.b

Estimate the acceleration of the bottle when it is at its maximum height.

[2]

Markscheme

Attempt to calculate gradient of line at t = 1.2 s ✓

«−» 9.8 «m s⁻²» (accept 9.6 − 10.0) ✓

c.

[Maximum mark: 2]

23M.2.SL.TZ2.c

Some water from the beaker is accidentally transferred with the cube.

Discuss how this will affect the value of the calculated specific heat capacity of the cube.

[2]

Markscheme

cube specific heat will be too large/increased value/overestimate ✓

additional «thermal» energy transferred

OR

temperature rise of water will be larger

OR

temperature drop of cube will be smaller ✓

d.

[Maximum mark: 2]

23M.2.SL.TZ2.d

The maximum height reached by the bottle is greater with an air–water mixture than with only high-pressure air in the bottle.

Assume that the speed at which the propellant leaves the bottle is the same in both cases.

Explain why the bottle reaches a greater maximum height with an air–water mixture.

[2]

Markscheme

Mass «leaving the bottle per second» will be larger for air–water ✓

the momentum change/force is greater ✓

Allow opposite argument for air only

e.

[Maximum mark: 2]

23M.2.SL.TZ2.e

A beta-minus particle and an alpha particle have the same initial kinetic energy.

Outline why the beta-minus particle can travel further in air than the alpha particle.

[2]

Markscheme

Alphas have double charge «and so are better ionisers »✓

alphas have more mass and therefore slower «for same energy» ✓

so longer time/more likely to interact with the «atomic» electrons/atoms «and therefore better ionisers» ✓

Accept reverse argument in terms of betas travelling faster.

i.

[Maximum mark: 1]

23M.2.SL.TZ2.i

outline the magnetic force acting on it due to the current in PQ.

[1]

Markscheme

«The force is» away from PQ/repulsive/to the right ✓

ii.

[Maximum mark: 1]

23M.2.SL.TZ2.ii

proton number.

[1]

Markscheme

2 ✓

iii.

[Maximum mark: 2]

23M.2.SL.TZ2.iii

Determine the ratio $\frac{luminosity of star X}{luminosity of star Y}$ .

[2]

Markscheme

$\frac{Luminosity of Star X}{Luminosity of Star Y}$ = $\frac{b_{x} d_{x}^{2}}{b_{y} d_{y}^{2}}$ ✓

= 10.8 ≈ 11 ✓

Award MP1 if ratio shown with distance or parallax angle.

Award MP1 for any correct substitution into ratio expression

Award [2] for BCA

Allow ECF for incorrect distances from b(i) or b(ii).

1.

[Maximum mark: 20]

EXE.2.HL.TZ0.1

A nuclear power station uses uranium-235 ( ${}_{92}^{235}U$ ) as fuel. One possible fission reaction of ${}_{92}^{235}U$ is

${}_{92}^{235}U + {}_{0}^{1}n \to {}_{52}^{132}{Te} + {}_{40}^{101}{Zr} + 3 {}_{0}^{1}n$

(a.i)

State the principal energy change in nuclear fission.

[1]

Markscheme

Mass-energy «of uranium» into kinetic energy of fission products ✓

(a.ii)

The energy released in the reaction is about 180 MeV. Estimate, in J, the energy released when 1 kg of ${}_{92}^{235}U$ undergoes fission.

[3]

Markscheme

Mass of uranium nucleus $\approx 235 u$ ✓

$\frac{energy}{mass} = \frac{180 \times 10^{6} \times 1.60 \times 10^{- 19}}{235 \times 1.661 \times 10^{- 27}}$ ✓

$7.4 \times 10^{13}$ «J» ✓

One of the products of the reaction is a nucleus of tellurium-132 ( ${}_{52}^{132}{Te}$ ). The diagram shows the location of ${}_{52}^{132}{Te}$ in a table of nuclides in which the proton number of a nuclide is plotted against its neutron number. The nuclides shown in black are stable.

(b.i)

State and explain the decay mode of ${}_{52}^{132}{Te}$ .

[2]

Markscheme

beta minus decay ✓

${}_{52}^{132}{Te}$ has more neutrons / higher $\frac{N}{Z}$ ratio than stable nuclides of similar A «and beta minus reduces $\frac{N}{Z}$ » ✓

A sample of pure ${}_{52}^{132}{Te}$ is extracted from some spent nuclear fuel from the reactor. The graph shows how the natural logarithm of the activity A of the sample varies with time t.

(b.ii)

Calculate, in s⁻¹, the initial activity of the sample.

[1]

Markscheme

$e^{25} = 7.2 \times 10^{10}$ «s⁻¹» ✓

(b.iii)

Show that the decay constant of a nuclide is given by −m, where m is the slope of the graph of lnA against t.

[1]

Markscheme

Takes ln of both sides of $A = A_{0} e^{- λ t}$ , leading to $\ln A = \ln A_{0} - λ t$ ✓

«hence slope $= - λ$ » ✓

(b.iv)

Determine, in days, the half-life of ${}_{52}^{132}{Te}$ .

[2]

Markscheme

Slope =«−» $2.5 \times 10^{- 6}$ «s⁻¹» ✓

$T_{\frac{1}{2}} =$ « $\frac{\ln 2}{2.5 \times 10^{- 6} \times 60 \times 60 \times 24} =$ » 3.2 «days» ✓

The nuclear power station uses high-pressure gas to power an electrical generator. The gas circulates between the heat exchanger and the turbine of the generator.

(c.i)

Outline the role of the heat exchanger in a nuclear power station.

[1]

Markscheme

Collects thermal energy from the coolant and delivers it to the gas ✓

Prevents the «irradiated» coolant from leaving the reactor vessel ✓

The working gas of the turbine undergoes a cyclic change that can be modelled as the cycle ABCDA shown in the pressure-volume diagram.

The cycle consists of an isobaric expansion AB, adiabatic expansion BC, isobaric compression CD and adiabatic compression DA. The cycle is drawn for a quantity of 1.0 mol of monatomic ideal gas.

(c.ii)

Calculate the maximum temperature of the gas during the cycle.

[3]

Markscheme

Correct read offs of pressure and volume at B ✓

$8.0 \times 10^{6} \times 1.6 \times 10^{- 3} = 1.0 \times 8.31 \times T$ ✓

$T = 1500$ «K» ✓

The following data are given about the work W done by the gas and thermal energy Q transferred to the gas during each change:

Change	W / kJ	Q / kJ
AB	8.23	20.58
BC	9.11	0
CD	−4.32	−10.81
DA	−3.25	0

(c.iii)

Outline why the entropy of the gas remains constant during changes BC and DA.

[1]

Markscheme

From $Δ S = \frac{Δ Q}{T}$ , the change in entropy is zero when $Δ Q = 0$ ✓

(c.iv)

Determine the efficiency of the cycle.

[2]

Markscheme

Net work done = « $8.23 + 9.11 - 4.32 - 3.25$ » 9.77 «kJ» ✓

Efficiency =« $\frac{9.77}{20.58} =$ » 0.47 ✓

During a maintenance shutdown of the reactor, the gas supply to the turbine is cut off and the turbine gradually comes to rest. The diagram shows how the angular speed of the turbine varies with time t.

(d)

Show that the rotational kinetic energy of the turbine decreases at a constant rate.

[3]

Markscheme

Rotational KE is proportional to $ω^{2}$ ✓

Calculation of $ω^{2}$ for at least four points, e.g. {96.1, 76.7, 57.6, 38.4, 19.3}×10³

Shows that the differences in equal time intervals are approximately the same, e.g. {19.4, 19.1, 19.2, 19.1, 19.3}×10³✓

Allow a tolerance of ±1×10³s⁻² from the values stated in MP2.

10.

[Maximum mark: 7]

EXE.2.HL.TZ0.10

The graph shows the variation in torque with time applied to a rotating flywheel.

(a)

Calculate the angular impulse applied to the flywheel.

[2]

Markscheme

Attempt to find area of triangle ✓

$\frac{1}{2} \times 8.0 \times 180 = 720$ «kg m² s⁻¹» ✓

(b)

The angular speed of the flywheel increased by 280 rad s⁻¹ during the application of the angular impulse.

Determine the moment of inertia of the flywheel.

[2]

Markscheme

Use of $Δ L = Δ (I ω)$ ✓

2.6 kg m² ✓

(c)

The flywheel was rotating at 150 rev per minute before the application of the angular impulse. Determine the change in angular rotational energy of the flywheel during the application of the flywheel.

[3]

Markscheme

Correct conversions to a consistent set of $ω$ units ✓

$\frac{1}{2} \times L \times (ω_{f}^{2} - ω_{i}^{2})$ or correct substitution seen ✓

113 kJ ✓

a.

[Maximum mark: 2]

EXE.2.HL.TZ0.10a

(a)

Calculate the angular impulse applied to the flywheel.

[2]

Markscheme

Attempt to find area of triangle ✓

$\frac{1}{2} \times 8.0 \times 180 = 720$ «kg m² s⁻¹» ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.10b

(b)

The angular speed of the flywheel increased by 280 rad s⁻¹ during the application of the angular impulse.

Determine the moment of inertia of the flywheel.

[2]

Markscheme

Use of $Δ L = Δ (I ω)$ ✓

2.6 kg m² ✓

c.

[Maximum mark: 3]

EXE.2.HL.TZ0.10c

(c)

The flywheel was rotating at 150 rev per minute before the application of the angular impulse. Determine the change in angular rotational energy of the flywheel during the application of the flywheel.

[3]

Markscheme

Correct conversions to a consistent set of $ω$ units ✓

$\frac{1}{2} \times L \times (ω_{f}^{2} - ω_{i}^{2})$ or correct substitution seen ✓

113 kJ ✓

11.

[Maximum mark: 6]

EXE.2.HL.TZ0.11

The particles of an ideal gas initially occupy one half of an isolated container, whose second half is initially empty. The gas is then allowed to expand freely into the second half. The diagram shows two configurations of the gas: the initial configuration A and configuration B, in which equal numbers of particles occupy each half of the container.

When a particle moves to a new position within the same half of the container, the microstate of the gas is considered unchanged. When a particle moves to the other half of the container, a new microstate is formed.

(a)

Explain why the gas in configuration B has a greater number of microstates than in A.

[3]

Markscheme

Configuration A has only one microstate ✓

In configuration B, pairs of particles can be swapped between the halves ✓

Every such change gives rise to a new microstate «so there is a large number of microstates in B» ✓

(b)

Deduce, with reference to entropy, that the expansion of the gas from the initial configuration A is irreversible.

[3]

Markscheme

The entropy of the gas is related to the number of microstates

OR

$S_{A} = k_{B} \ln Ω_{A}$ and $S_{A} = k_{B} \ln Ω_{B}$ ✓

Since $Ω_{B} > Ω_{A}$ , the entropy in configuration B is greater ✓

A process that results in an increase of entropy in an isolated system is irreversible ✓

a.

[Maximum mark: 3]

EXE.2.HL.TZ0.11a

(a)

Explain why the gas in configuration B has a greater number of microstates than in A.

[3]

Markscheme

Configuration A has only one microstate ✓

In configuration B, pairs of particles can be swapped between the halves ✓

Every such change gives rise to a new microstate «so there is a large number of microstates in B» ✓

b.

[Maximum mark: 3]

EXE.2.HL.TZ0.11b

(b)

Deduce, with reference to entropy, that the expansion of the gas from the initial configuration A is irreversible.

[3]

Markscheme

The entropy of the gas is related to the number of microstates

OR

$S_{A} = k_{B} \ln Ω_{A}$ and $S_{A} = k_{B} \ln Ω_{B}$ ✓

Since $Ω_{B} > Ω_{A}$ , the entropy in configuration B is greater ✓

A process that results in an increase of entropy in an isolated system is irreversible ✓

12.

[Maximum mark: 8]

EXE.2.HL.TZ0.12

An isolated system consists of six particles. The total energy of the system is 6E, where E is a constant. The particles can randomly exchange energy between one another, in integer multiples of E.

(a)

State what is meant by an isolated system.

[1]

Markscheme

Neither mass nor energy is exchanged with the surroundings ✓

The energy diagram shows two possible configurations of the system. Each dot in the diagram represents one particle. In configuration A, one particle has energy 6E and the remaining particles have zero energy. In configuration B, three particles have energies 3E, 2E and E, and the remaining particles have zero energy.

(b)

State and explain the number of microstates of the system in configuration A.

[2]

Markscheme

6 microstates ✓

Any of the six particles can be the one of the highest energy ✓

(c)

Configuration B has 120 microstates. Calculate the entropy difference between configurations B and A. State the answer in terms of $k_{B}$ .

[2]

Markscheme

$k_{B} \ln 120 - k_{B} \ln 6$ ✓

$3.0 k_{B}$ ✓

(d)

The system is initially in configuration A. Comment, with reference to the second law of thermodynamics and your answer in (c), on the likely evolution of the system.

[3]

Markscheme

The second law predicts that isolated systems spontaneously evolve towards high-entropy states ✓

From (c), the entropy of B is greater than that of A ✓

The final state will likely be similar to B / contain relatively many low-energy particles «of different energies» ✓

a.

[Maximum mark: 1]

EXE.2.HL.TZ0.12a

(a)

State what is meant by an isolated system.

[1]

Markscheme

Neither mass nor energy is exchanged with the surroundings ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.12b

The energy diagram shows two possible configurations of the system. Each dot in the diagram represents one particle. In configuration A, one particle has energy 6E and the remaining particles have zero energy. In configuration B, three particles have energies 3E, 2E and E, and the remaining particles have zero energy.

(b)

State and explain the number of microstates of the system in configuration A.

[2]

Markscheme

6 microstates ✓

Any of the six particles can be the one of the highest energy ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.12c

(c)

Configuration B has 120 microstates. Calculate the entropy difference between configurations B and A. State the answer in terms of $k_{B}$ .

[2]

Markscheme

$k_{B} \ln 120 - k_{B} \ln 6$ ✓

$3.0 k_{B}$ ✓

d.

[Maximum mark: 3]

EXE.2.HL.TZ0.12d

(d)

The system is initially in configuration A. Comment, with reference to the second law of thermodynamics and your answer in (c), on the likely evolution of the system.

[3]

Markscheme

The second law predicts that isolated systems spontaneously evolve towards high-entropy states ✓

From (c), the entropy of B is greater than that of A ✓

The final state will likely be similar to B / contain relatively many low-energy particles «of different energies» ✓

a.

[Maximum mark: 3]

EXE.2.HL.TZ0.13a

(a)

Outline, using these two cases as examples, the distinction between a microstate and a macrostate.

[3]

Markscheme

Idea that a microstate is one (unique) arrangement of a system assuming that each entity in the system is distinguishable OWTTE ✓

Idea that a macrostate is an arrangement of microstates which have an identical outcome when individuals are not treated as distinguishable ✓

Suitable link to this example: e.g. there is only one arrangement (macrostate) in which there are 49 molecules in one container even though there are 50 (microstates or) ways to arrange this ✓

14.

[Maximum mark: 7]

EXE.2.HL.TZ0.14

The table shows some of the macrostates and microstates for 10 identical coins tossed at random that can land either heads or tails upwards.

Macrostate		Number of microstates
heads	tails
10	0	1
9	1	10
8	2	45
7	3	120
6	4	210
5	5	252

There are a total of 1024 microstates for this system.

(a)

Determine the fractional number of throws for which the three most likely macrostates occur.

[3]

Markscheme

Recognises that 4 heads and 6 tails also required ✓

Total number of microstates = 672 ✓

Fractional number = 672/1024 = 0.66 ✓

Allow ecf for MP3

(b)

A throw is made once every minute. Estimate the average time required before a throw occurs where all coins are heads or all coins are tails.

[2]

Markscheme

Two chances in 1024 so once every 512 throws ✓

512 throws take 8.5 h so (a reasonable estimate is half way through) on average 4.3 h ✓

(c)

In one throw the coins all land heads upwards. The following throw results in 7 heads and 3 tails. Calculate, in terms of $k_{B}$ , the change in entropy between the two throws.

[2]

Markscheme

$Δ S = k_{B} (\ln 120 - \ln 1)$ ✓

$= 4.8 k_{B}$ ✓

a.

[Maximum mark: 3]

EXE.2.HL.TZ0.14a

(a)

Determine the fractional number of throws for which the three most likely macrostates occur.

[3]

Markscheme

Recognises that 4 heads and 6 tails also required ✓

Total number of microstates = 672 ✓

Fractional number = 672/1024 = 0.66 ✓

Allow ecf for MP3

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.14b

(b)

A throw is made once every minute. Estimate the average time required before a throw occurs where all coins are heads or all coins are tails.

[2]

Markscheme

Two chances in 1024 so once every 512 throws ✓

512 throws take 8.5 h so (a reasonable estimate is half way through) on average 4.3 h ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.14c

(c)

In one throw the coins all land heads upwards. The following throw results in 7 heads and 3 tails. Calculate, in terms of $k_{B}$ , the change in entropy between the two throws.

[2]

Markscheme

$Δ S = k_{B} (\ln 120 - \ln 1)$ ✓

$= 4.8 k_{B}$ ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.15ai

(i)

orbital speed;

[1]

Markscheme

« $\sqrt{\frac{6.67 \times 10^{- 11} \times 5.97 \times 10^{24}}{6.70 \times 10^{6}}} =$ » $7.71 \times 10^{3}$ «m s⁻¹» ✓

ii.

[Maximum mark: 1]

EXE.2.HL.TZ0.15aii

(ii)

escape speed from its orbit.

[1]

Markscheme

« $\sqrt{\frac{2 \times 6.67 \times 10^{- 11} \times 5.97 \times 10^{24}}{6.70 \times 10^{6}}}$ = » $1.09 \times 10^{4} « m s^{- 1} »$ ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.15ci

(i)

in its initial circular orbit;

[1]

Markscheme

Negative ✓

ii.

[Maximum mark: 1]

EXE.2.HL.TZ0.15cii

(ii)

in its final orbit.

[1]

Markscheme

Positive ✓

16.

[Maximum mark: 7]

EXE.2.HL.TZ0.16

(a)

The radius of the dwarf planet Pluto is 1.19 x 10⁶ m. The acceleration due to gravity at its surface is 0.617 m s⁻².

Determine the escape speed for an object at the surface of Pluto.

[4]

Markscheme

$v_{esc} = \sqrt{\frac{2 G M}{r}}$ AND $g = \frac{G M}{r^{2}}$ seen ✓

$v_{esc} = \sqrt{\frac{2 g r^{2}}{r}}$ ✓

Leading to $v_{esc} = \sqrt{2 g r}$ ✓

1.2 km s⁻¹✓

(b)

Pluto rotates about an axis through its centre. Its rotation is in the opposite sense to that of the Earth, i.e. from east to west.

Explain the advantage of an object launching from the equator of Pluto and travelling to the west.

[3]

Markscheme

Object at equator has the maximum linear/tangential speed possible ✓

It therefore has maximum kinetic energy before takeoff (and this is not required from the fuel) ✓

Idea that the object is already moving in direction of planet before takeoff ✓

a.

[Maximum mark: 4]

EXE.2.HL.TZ0.16a

(a)

The radius of the dwarf planet Pluto is 1.19 x 10⁶ m. The acceleration due to gravity at its surface is 0.617 m s⁻².

Determine the escape speed for an object at the surface of Pluto.

[4]

Markscheme

$v_{esc} = \sqrt{\frac{2 G M}{r}}$ AND $g = \frac{G M}{r^{2}}$ seen ✓

$v_{esc} = \sqrt{\frac{2 g r^{2}}{r}}$ ✓

Leading to $v_{esc} = \sqrt{2 g r}$ ✓

1.2 km s⁻¹✓

b.

[Maximum mark: 3]

EXE.2.HL.TZ0.16b

(b)

Pluto rotates about an axis through its centre. Its rotation is in the opposite sense to that of the Earth, i.e. from east to west.

Explain the advantage of an object launching from the equator of Pluto and travelling to the west.

[3]

Markscheme

Object at equator has the maximum linear/tangential speed possible ✓

It therefore has maximum kinetic energy before takeoff (and this is not required from the fuel) ✓

Idea that the object is already moving in direction of planet before takeoff ✓

17.

[Maximum mark: 7]

EXE.2.HL.TZ0.17

In a Compton scattering experiment, an X-ray photon interacts with a free, stationary electron. The electron recoils with an energy of 500 eV. The wavelength of the scattered photon is $7.80 \times 10^{- 11}$ m.

(a)

Show that the energy of the scattered photon is about 16 keV.

[1]

Markscheme

$\frac{1.24 \times 10^{- 6}}{7.80 \times 10^{- 11}}$ «= 15.9 keV» ✓

(b)

Determine the wavelength of the incident photon.

[2]

Markscheme

Energy of incident photon =« $15.9 \times 10^{3} + 500 =$ » $16.4 \times 10^{3}$ «eV» ✓

Wavelength of incident photon =« $\frac{1.24 \times 10^{- 6}}{16.4 \times 10^{3}} =$ » $7.56 \times 10^{- 11}$ «m» ✓

(c)

Outline why the results of the experiment are inconsistent with the wave model of electromagnetic radiation.

[2]

Markscheme

The wavelength of the X-rays changes ✓

According to the wave model, the wavelength of the incident and scattered X-rays should be the same ✓

(d)

Calculate the scattering angle of the photon.

[2]

Markscheme

$(7.80 - 7.56) \times 10^{- 11} = \frac{1.24 \times 10^{- 6}}{0.511 \times 10^{6}} (1 - \cos θ)$ ✓

$θ = 89 °$ ✓

a.

[Maximum mark: 1]

EXE.2.HL.TZ0.17a

(a)

Show that the energy of the scattered photon is about 16 keV.

[1]

Markscheme

$\frac{1.24 \times 10^{- 6}}{7.80 \times 10^{- 11}}$ «= 15.9 keV» ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.17b

(b)

Determine the wavelength of the incident photon.

[2]

Markscheme

Energy of incident photon =« $15.9 \times 10^{3} + 500 =$ » $16.4 \times 10^{3}$ «eV» ✓

Wavelength of incident photon =« $\frac{1.24 \times 10^{- 6}}{16.4 \times 10^{3}} =$ » $7.56 \times 10^{- 11}$ «m» ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.17c

(c)

Outline why the results of the experiment are inconsistent with the wave model of electromagnetic radiation.

[2]

Markscheme

The wavelength of the X-rays changes ✓

According to the wave model, the wavelength of the incident and scattered X-rays should be the same ✓

d.

[Maximum mark: 2]

EXE.2.HL.TZ0.17d

(d)

Calculate the scattering angle of the photon.

[2]

Markscheme

$(7.80 - 7.56) \times 10^{- 11} = \frac{1.24 \times 10^{- 6}}{0.511 \times 10^{6}} (1 - \cos θ)$ ✓

$θ = 89 °$ ✓

18.

[Maximum mark: 8]

EXE.2.HL.TZ0.18

A beam of electrons is incident on a thin crystalline sample of graphite. The electrons emerging from the sample form a pattern on a fluorescent screen. The pattern consists of a series of bright and dark rings concentric with the direction of the incident beam.

(a)

Outline why the pattern observed on the screen is an evidence for matter waves.

[2]

Markscheme

The pattern is formed when the electrons scattered from adjacent planes in the graphite crystal undergo interference / diffract ✓

Interference / diffraction is a property of waves only ✓

The beam is produced by accelerating electrons through an electric potential difference U.

(b)

A typical interatomic distance in the graphite crystal is of the order of $2 \times 10^{- 10}$ m. Estimate the minimum value of U for the pattern in (a) to be formed on the screen.

[4]

Markscheme

The de Broglie wavelength of the electrons should be comparable to or shorter than the interatomic distance / $2 \times 10^{- 10}$ m ✓

Momentum of electrons $\approx \frac{6.63 \times 10^{- 34}}{2 \times 10^{- 10}}$ « $= 3.32 \times 10^{- 2}$ N s» ✓

Kinetic energy of electrons = « $\frac{p^{2}}{2 m} = \frac{{(3.32 \times 10^{- 24})}^{2}}{2 \times 9.11 \times 10^{- 31}} =$ » $6.0 \times 10^{- 18}$ «J» ✓

U = « $\frac{E}{e} = \frac{6.10 \times 10^{- 18}}{1.6 \times 10^{- 19}} =$ »40 «V» ✓

(c)

Protons can also be accelerated by the same potential difference U. Compare, without calculation, the de Broglie wavelength of the protons to that of the electrons.

[2]

Markscheme

The protons have the same energy but greater mass hence a greater momentum than the electrons ✓

From $λ = \frac{h}{p}$ , the protons will have a shorter wavelength ✓

a.

[Maximum mark: 2]

EXE.2.HL.TZ0.18a

(a)

Outline why the pattern observed on the screen is an evidence for matter waves.

[2]

Markscheme

The pattern is formed when the electrons scattered from adjacent planes in the graphite crystal undergo interference / diffract ✓

Interference / diffraction is a property of waves only ✓

b.

[Maximum mark: 4]

EXE.2.HL.TZ0.18b

The beam is produced by accelerating electrons through an electric potential difference U.

(b)

A typical interatomic distance in the graphite crystal is of the order of $2 \times 10^{- 10}$ m. Estimate the minimum value of U for the pattern in (a) to be formed on the screen.

[4]

Markscheme

The de Broglie wavelength of the electrons should be comparable to or shorter than the interatomic distance / $2 \times 10^{- 10}$ m ✓

Momentum of electrons $\approx \frac{6.63 \times 10^{- 34}}{2 \times 10^{- 10}}$ « $= 3.32 \times 10^{- 2}$ N s» ✓

Kinetic energy of electrons = « $\frac{p^{2}}{2 m} = \frac{{(3.32 \times 10^{- 24})}^{2}}{2 \times 9.11 \times 10^{- 31}} =$ » $6.0 \times 10^{- 18}$ «J» ✓

U = « $\frac{E}{e} = \frac{6.10 \times 10^{- 18}}{1.6 \times 10^{- 19}} =$ »40 «V» ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.18c

(c)

Protons can also be accelerated by the same potential difference U. Compare, without calculation, the de Broglie wavelength of the protons to that of the electrons.

[2]

Markscheme

The protons have the same energy but greater mass hence a greater momentum than the electrons ✓

From $λ = \frac{h}{p}$ , the protons will have a shorter wavelength ✓

19.

[Maximum mark: 8]

EXE.2.HL.TZ0.19

(a)

State the de Broglie hypothesis.

[2]

Markscheme

a moving particle has wave properties ✓

de Broglie wavelength = Planck constant÷momentum (must define p if quoted as equation) ✓

A beam of electrons is accelerated from rest through a potential difference of 500 V maintained between two plates in a vacuum. The electrons then travel through a circular hole in the +500 V plate.

(b)

Calculate the maximum speed of the electrons in the beam.

[2]

Markscheme

use of ½ mv² = eV ✓

1.33 × 10⁷ m s⁻¹✓

(c)

After passing through the circular hole the electrons strike a fluorescent screen.

Predict whether an apparatus such as this can demonstrate that moving electrons have wave properties.

[4]

Markscheme

idea that de Broglie wavelength and hole size must be similar ✓

Use of $λ = \frac{h}{\sqrt{2 m e V}}$ ✓

Leading to $λ$ around 5 x 10⁻¹¹m (which is unrealistic for a practical situation) ✓

It would not be possible to construct this hole
OR
hole must be smaller than an atom so impossible ✓

a.

[Maximum mark: 2]

EXE.2.HL.TZ0.19a

(a)

State the de Broglie hypothesis.

[2]

Markscheme

a moving particle has wave properties ✓

de Broglie wavelength = Planck constant÷momentum (must define p if quoted as equation) ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.19b

A beam of electrons is accelerated from rest through a potential difference of 500 V maintained between two plates in a vacuum. The electrons then travel through a circular hole in the +500 V plate.

(b)

Calculate the maximum speed of the electrons in the beam.

[2]

Markscheme

use of ½ mv² = eV ✓

1.33 × 10⁷ m s⁻¹✓

c.

[Maximum mark: 4]

EXE.2.HL.TZ0.19c

(c)

After passing through the circular hole the electrons strike a fluorescent screen.

Predict whether an apparatus such as this can demonstrate that moving electrons have wave properties.

[4]

Markscheme

idea that de Broglie wavelength and hole size must be similar ✓

Use of $λ = \frac{h}{\sqrt{2 m e V}}$ ✓

Leading to $λ$ around 5 x 10⁻¹¹m (which is unrealistic for a practical situation) ✓

It would not be possible to construct this hole
OR
hole must be smaller than an atom so impossible ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.1ai

(i)

State the principal energy change in nuclear fission.

[1]

Markscheme

Mass-energy «of uranium» into kinetic energy of fission products ✓

ii.

[Maximum mark: 3]

EXE.2.HL.TZ0.1aii

(ii)

The energy released in the reaction is about 180 MeV. Estimate, in J, the energy released when 1 kg of ${}_{92}^{235}U$ undergoes fission.

[3]

Markscheme

Mass of uranium nucleus $\approx 235 u$ ✓

$\frac{energy}{mass} = \frac{180 \times 10^{6} \times 1.60 \times 10^{- 19}}{235 \times 1.661 \times 10^{- 27}}$ ✓

$7.4 \times 10^{13}$ «J» ✓

i.

[Maximum mark: 2]

EXE.2.HL.TZ0.1bi

(i)

State and explain the decay mode of ${}_{52}^{132}{Te}$ .

[2]

Markscheme

beta minus decay ✓

${}_{52}^{132}{Te}$ has more neutrons / higher $\frac{N}{Z}$ ratio than stable nuclides of similar A «and beta minus reduces $\frac{N}{Z}$ » ✓

ii.

[Maximum mark: 1]

EXE.2.HL.TZ0.1bii

A sample of pure ${}_{52}^{132}{Te}$ is extracted from some spent nuclear fuel from the reactor. The graph shows how the natural logarithm of the activity A of the sample varies with time t.

(ii)

Calculate, in s⁻¹, the initial activity of the sample.

[1]

Markscheme

$e^{25} = 7.2 \times 10^{10}$ «s⁻¹» ✓

iii.

[Maximum mark: 1]

EXE.2.HL.TZ0.1biii

(iii)

Show that the decay constant of a nuclide is given by −m, where m is the slope of the graph of lnA against t.

[1]

Markscheme

Takes ln of both sides of $A = A_{0} e^{- λ t}$ , leading to $\ln A = \ln A_{0} - λ t$ ✓

«hence slope $= - λ$ » ✓

iv.

[Maximum mark: 2]

EXE.2.HL.TZ0.1biv

(iv)

Determine, in days, the half-life of ${}_{52}^{132}{Te}$ .

[2]

Markscheme

Slope =«−» $2.5 \times 10^{- 6}$ «s⁻¹» ✓

$T_{\frac{1}{2}} =$ « $\frac{\ln 2}{2.5 \times 10^{- 6} \times 60 \times 60 \times 24} =$ » 3.2 «days» ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.1ci

(i)

Outline the role of the heat exchanger in a nuclear power station.

[1]

Markscheme

Collects thermal energy from the coolant and delivers it to the gas ✓

Prevents the «irradiated» coolant from leaving the reactor vessel ✓

ii.

[Maximum mark: 3]

EXE.2.HL.TZ0.1cii

The working gas of the turbine undergoes a cyclic change that can be modelled as the cycle ABCDA shown in the pressure-volume diagram.

The cycle consists of an isobaric expansion AB, adiabatic expansion BC, isobaric compression CD and adiabatic compression DA. The cycle is drawn for a quantity of 1.0 mol of monatomic ideal gas.

(ii)

Calculate the maximum temperature of the gas during the cycle.

[3]

Markscheme

Correct read offs of pressure and volume at B ✓

$8.0 \times 10^{6} \times 1.6 \times 10^{- 3} = 1.0 \times 8.31 \times T$ ✓

$T = 1500$ «K» ✓

iii.

[Maximum mark: 1]

EXE.2.HL.TZ0.1ciii

The following data are given about the work W done by the gas and thermal energy Q transferred to the gas during each change:

Change	W / kJ	Q / kJ
AB	8.23	20.58
BC	9.11	0
CD	−4.32	−10.81
DA	−3.25	0

(iii)

Outline why the entropy of the gas remains constant during changes BC and DA.

[1]

Markscheme

From $Δ S = \frac{Δ Q}{T}$ , the change in entropy is zero when $Δ Q = 0$ ✓

iv.

[Maximum mark: 2]

EXE.2.HL.TZ0.1civ

(iv)

Determine the efficiency of the cycle.

[2]

Markscheme

Net work done = « $8.23 + 9.11 - 4.32 - 3.25$ » 9.77 «kJ» ✓

Efficiency =« $\frac{9.77}{20.58} =$ » 0.47 ✓

d.

[Maximum mark: 3]

EXE.2.HL.TZ0.1d

During a maintenance shutdown of the reactor, the gas supply to the turbine is cut off and the turbine gradually comes to rest. The diagram shows how the angular speed of the turbine varies with time t.

(d)

Show that the rotational kinetic energy of the turbine decreases at a constant rate.

[3]

Markscheme

Rotational KE is proportional to $ω^{2}$ ✓

Calculation of $ω^{2}$ for at least four points, e.g. {96.1, 76.7, 57.6, 38.4, 19.3}×10³

Shows that the differences in equal time intervals are approximately the same, e.g. {19.4, 19.1, 19.2, 19.1, 19.3}×10³✓

Allow a tolerance of ±1×10³s⁻² from the values stated in MP2.

2.

[Maximum mark: 20]

EXE.2.HL.TZ0.2

A geophone is an instrument designed to measure the movement of ground rocks.

When the ground moves, the magnet-spring system oscillates relative to the coil. An emf is generated in the coil. The magnitude of this emf is proportional to the speed of the magnet relative to the coil.

(a.i)

State the movement direction for which the geophone has its greatest sensitivity.

[1]

Markscheme

Vertical direction / parallel to springs ✓

(a.ii)

Outline how an emf is generated in the coil.

[2]

Markscheme

The magnetic field moves relative to the coil ✓

As field lines cut the coil, forces act on (initially stationary) electrons in the wire (and these move producing an emf) ✓

(a.iii)

Explain why the magnitude of the emf is related to the amplitude of the ground movement.

[3]

Markscheme

The springs have a natural time period for the oscillation ✓

A greater amplitude of movement leads to higher magnet speed (with constant time period) ✓

So field lines cut coil more quickly leading to greater emf ✓

(a.iv)

In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.

Calculate the rate of flux change that leads to this emf.

[2]

Markscheme

Use of $ε = - \frac{Δ (N Φ)}{Δ t}$ ✓

$\frac{ΔΦ}{Δ t} = (-) \frac{65 \times 10^{- 3}}{150}$

$= 0.43$ mWb s⁻¹ ✓

(b)

Suggest two changes to the system that will make the geophone more sensitive.

[4]

Markscheme

Any two suggestions from:

Increase number of turns in coil ✓
Because more flux cutting per cycle ✓

Increase field strength of magnet ✓
So that there are more field lines ✓

Change mass-spring system so that time period decreases ✓
So magnet will be moving faster for given amplitude of movement ✓

The geophone is mounted on the ground at point Z and an explosion is produced at point W some distance away. Sound from the explosion travels to the geophone via the clay layer in the ground.

Diagram not to scale

The speed of sound in clay is 3.00 km s⁻¹; the speed of sound in sandstone is 4.70 km s⁻¹

(c.i)

Show that, when sound travels from clay to sandstone, the critical angle is approximately 40°.

[2]

Markscheme

_cn_s $= \frac{3.0}{4.7} = 1.57$ ✓

Critical angle $= \sin^{- 1} (\frac{1}{1.57}) = 39.6 °$ ✓

(c.ii)

The angle between the clay–air surface and path 1 is 80°.

Draw, on the diagram, the subsequent path of a sound wave that travels initially in the clay along path 1.

[2]

Markscheme

ray shown reflected back into the clay (and then to Z) at (by eye) the incidence angle ✓

ray shown refracted into the sandstone with angle of refraction greater than angle of incidence (by eye) ✓

Another explosion is produced at X. The sound from this explosion is detected twice at the geophone at Z. Some sound travels directly from X to Z through clay along path 2. Other sound travels through clay via Y along path 3.

The vertical thickness of the clay layer is d. The distance XZ is 80.0 m.

The time between the arrival of the sounds due to the path difference is 6.67 ms.

(d)

Calculate d.

[4]

Markscheme

distance difference $= (3000 \times 0.00667 =) 19.8$ m ✓

½ distance difference $= 9.9$ m so YZ $= 49.9$ m ✓

$d = (\sqrt{{YZ}^{2} - {(\frac{XZ}{2})}^{2}}) = \sqrt{49 . 9^{2} - 40^{2}}$ ✓

29.8 m ✓

OR

Recognises situation as (almost) 3:4:5 triangle ✓

30 m (1 sf answer only accepted in this route) ✓

20.

[Maximum mark: 8]

EXE.2.HL.TZ0.20

(a)

The quantity $\frac{h}{m_{e} c}$ is known as the Compton wavelength.

Show that the Compton wavelength is about 2.4 pm.

[1]

Markscheme

2.43 pm ✓

(b)

A photon with a wavelength of 6.00 pm interacts with a stationary electron. After the interaction the photon's wavelength has changed by 2.43 pm.

(b.i)

State the wavelength of the photon after the interaction.

[1]

Markscheme

8.43 pm ✓

(b.ii)

Outline why the wavelength of the photon has changed.

[2]

Markscheme

(Energy of photon inversely prop to wavelength)

photon transfers some of its energy to the electron. ✓

If its energy decreases so its wavelength increases. ✓

(b.iii)

Deduce the scattering angle for the photon.

[2]

Markscheme

(for this interaction)

$Δ λ = \frac{h}{m_{e} c} (1 - \cos θ) = \frac{h}{m_{e} c}$ and therefore $(1 - \cos θ)$ must equal 1 ✓

so cos theta = 0 and theta = 90 deg ✓

(b.iv)

Determine, in J, the kinetic energy of the electron after the interaction.

[2]

Markscheme

(Because energy is conserved)

$E_{k} = h c (\frac{1}{λ_{i}} - \frac{1}{λ_{f}})$

$= 6.63 \times 10^{- 34} \times 3.00 \times 10^{8} (\frac{1}{6.00 \times 10^{- 12}} - \frac{1}{8.43 \times 10^{- 12}})$ ✓

9.6 fJ ✓

a.

[Maximum mark: 1]

EXE.2.HL.TZ0.20a

(a)

The quantity $\frac{h}{m_{e} c}$ is known as the Compton wavelength.

Show that the Compton wavelength is about 2.4 pm.

[1]

Markscheme

2.43 pm ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.20bi

(i)

State the wavelength of the photon after the interaction.

[1]

Markscheme

8.43 pm ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.20bii

(ii)

Outline why the wavelength of the photon has changed.

[2]

Markscheme

(Energy of photon inversely prop to wavelength)

photon transfers some of its energy to the electron. ✓

If its energy decreases so its wavelength increases. ✓

iii.

[Maximum mark: 2]

EXE.2.HL.TZ0.20biii

(iii)

Deduce the scattering angle for the photon.

[2]

Markscheme

(for this interaction)

$Δ λ = \frac{h}{m_{e} c} (1 - \cos θ) = \frac{h}{m_{e} c}$ and therefore $(1 - \cos θ)$ must equal 1 ✓

so cos theta = 0 and theta = 90 deg ✓

iv.

[Maximum mark: 2]

EXE.2.HL.TZ0.20biv

(iv)

Determine, in J, the kinetic energy of the electron after the interaction.

[2]

Markscheme

(Because energy is conserved)

$E_{k} = h c (\frac{1}{λ_{i}} - \frac{1}{λ_{f}})$

$= 6.63 \times 10^{- 34} \times 3.00 \times 10^{8} (\frac{1}{6.00 \times 10^{- 12}} - \frac{1}{8.43 \times 10^{- 12}})$ ✓

9.6 fJ ✓

21.

[Maximum mark: 9]

EXE.2.HL.TZ0.21

(a)

Suggest one problem that is faced in dealing with the waste from nuclear fission reactors. Go on to outline how this problem is overcome.

[2]

Markscheme

Waste is very hot …

… So has to be placed in cooling ponds to transfer the (thermal) energy away ✓

OR

Waste is very radioactive … ✓

… So has to be placed in cooling ponds to absorb this radiation

OR

… So has to be handled remotely

OR

… So has to be transported in crash resistant casings / stored on site ✓

OR

Waste will be radioactive for thousands of years … ✓

… So storage needs to be (eventually) in geologically stable areas ✓

(b)

Strontium-90 is a waste product from nuclear reactors that has a decay constant of 7.63 x 10⁻¹⁰ s⁻¹. Determine, in s, the time that it takes for the activity of strontium-90 to decay to 2% of its original activity.

[2]

Markscheme

$\ln (0.02) = - (7.63 \times 10^{- 10}) t$ or equivalent seen ✓

$t = 5.1$ Gs ✓

(c)

The decay of one Strontium-90 nucleus leads to an energy release of about 0.52 MeV. The decay product of Strontium-90 is Yttrium-90 which itself decays to stable Zirconium-90 with a decay constant of 3.0 x 10⁻⁶ s⁻¹. The energy released in the decay of one Yttrium-90 nucleus is 2.3 MeV.

(c.i)

Calculate the energy released when one mole of strontium-90 decays to 2% of its original activity forming the stable daughter product.

[3]

Markscheme

Idea that the Yttrium half life is much less than Strontium so can assume all Yttrium energy is included. ✓

$0.98 \times 6.02 \times 10^{23} \times (0.52 + 2.3) \times 10^{6}$ seen ✓

$(= 1.67 \times 1030 eV)$

Answer $\times 1.6 \times 10^{- 19} = 270$ GJ ✓

(c.ii)

Strontium-90 decays to Zirconium-90 via two successive beta emissions. Discuss whether all the energy released when strontium-90 decays to Zirconium-90 can be transferred to a thermal form.

[2]

Markscheme

(No)
(anti-)neutrinos are released in (both) decays ✓

Carrying away energy because they interact poorly with matter ✓

Ignore arguments relating to energy transferred to nucleus as this appears eventually as thermal energy.

a.

[Maximum mark: 2]

EXE.2.HL.TZ0.21a

(a)

Suggest one problem that is faced in dealing with the waste from nuclear fission reactors. Go on to outline how this problem is overcome.

[2]

Markscheme

Waste is very hot …

… So has to be placed in cooling ponds to transfer the (thermal) energy away ✓

OR

Waste is very radioactive … ✓

… So has to be placed in cooling ponds to absorb this radiation

OR

… So has to be handled remotely

OR

… So has to be transported in crash resistant casings / stored on site ✓

OR

Waste will be radioactive for thousands of years … ✓

… So storage needs to be (eventually) in geologically stable areas ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.21b

(b)

Strontium-90 is a waste product from nuclear reactors that has a decay constant of 7.63 x 10⁻¹⁰ s⁻¹. Determine, in s, the time that it takes for the activity of strontium-90 to decay to 2% of its original activity.

[2]

Markscheme

$\ln (0.02) = - (7.63 \times 10^{- 10}) t$ or equivalent seen ✓

$t = 5.1$ Gs ✓

i.

[Maximum mark: 3]

EXE.2.HL.TZ0.21ci

(i)

Calculate the energy released when one mole of strontium-90 decays to 2% of its original activity forming the stable daughter product.

[3]

Markscheme

Idea that the Yttrium half life is much less than Strontium so can assume all Yttrium energy is included. ✓

$0.98 \times 6.02 \times 10^{23} \times (0.52 + 2.3) \times 10^{6}$ seen ✓

$(= 1.67 \times 1030 eV)$

Answer $\times 1.6 \times 10^{- 19} = 270$ GJ ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.21cii

(ii)

Strontium-90 decays to Zirconium-90 via two successive beta emissions. Discuss whether all the energy released when strontium-90 decays to Zirconium-90 can be transferred to a thermal form.

[2]

Markscheme

(No)
(anti-)neutrinos are released in (both) decays ✓

Carrying away energy because they interact poorly with matter ✓

Ignore arguments relating to energy transferred to nucleus as this appears eventually as thermal energy.

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.2ai

(i)

State the movement direction for which the geophone has its greatest sensitivity.

[1]

Markscheme

Vertical direction / parallel to springs ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.2aii

(ii)

Outline how an emf is generated in the coil.

[2]

Markscheme

The magnetic field moves relative to the coil ✓

As field lines cut the coil, forces act on (initially stationary) electrons in the wire (and these move producing an emf) ✓

iii.

[Maximum mark: 3]

EXE.2.HL.TZ0.2aiii

(iii)

Explain why the magnitude of the emf is related to the amplitude of the ground movement.

[3]

Markscheme

The springs have a natural time period for the oscillation ✓

A greater amplitude of movement leads to higher magnet speed (with constant time period) ✓

So field lines cut coil more quickly leading to greater emf ✓

iv.

[Maximum mark: 2]

EXE.2.HL.TZ0.2aiv

(iv)

In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.

Calculate the rate of flux change that leads to this emf.

[2]

Markscheme

Use of $ε = - \frac{Δ (N Φ)}{Δ t}$ ✓

$\frac{ΔΦ}{Δ t} = (-) \frac{65 \times 10^{- 3}}{150}$

$= 0.43$ mWb s⁻¹ ✓

b.

[Maximum mark: 4]

EXE.2.HL.TZ0.2b

(b)

Suggest two changes to the system that will make the geophone more sensitive.

[4]

Markscheme

Any two suggestions from:

Increase number of turns in coil ✓
Because more flux cutting per cycle ✓

Increase field strength of magnet ✓
So that there are more field lines ✓

Change mass-spring system so that time period decreases ✓
So magnet will be moving faster for given amplitude of movement ✓

i.

[Maximum mark: 2]

EXE.2.HL.TZ0.2ci

(i)

Show that, when sound travels from clay to sandstone, the critical angle is approximately 40°.

[2]

Markscheme

_cn_s $= \frac{3.0}{4.7} = 1.57$ ✓

Critical angle $= \sin^{- 1} (\frac{1}{1.57}) = 39.6 °$ ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.2cii

(ii)

The angle between the clay–air surface and path 1 is 80°.

Draw, on the diagram, the subsequent path of a sound wave that travels initially in the clay along path 1.

[2]

Markscheme

ray shown reflected back into the clay (and then to Z) at (by eye) the incidence angle ✓

ray shown refracted into the sandstone with angle of refraction greater than angle of incidence (by eye) ✓

d.

[Maximum mark: 4]

EXE.2.HL.TZ0.2d

Another explosion is produced at X. The sound from this explosion is detected twice at the geophone at Z. Some sound travels directly from X to Z through clay along path 2. Other sound travels through clay via Y along path 3.

The vertical thickness of the clay layer is d. The distance XZ is 80.0 m.

The time between the arrival of the sounds due to the path difference is 6.67 ms.

(d)

Calculate d.

[4]

Markscheme

distance difference $= (3000 \times 0.00667 =) 19.8$ m ✓

½ distance difference $= 9.9$ m so YZ $= 49.9$ m ✓

$d = (\sqrt{{YZ}^{2} - {(\frac{XZ}{2})}^{2}}) = \sqrt{49 . 9^{2} - 40^{2}}$ ✓

29.8 m ✓

OR

Recognises situation as (almost) 3:4:5 triangle ✓

30 m (1 sf answer only accepted in this route) ✓

3.

[Maximum mark: 10]

EXE.2.HL.TZ0.3

Ball A, moving in a horizontal direction at an initial speed of 2.0 m s⁻¹ collides with a stationary ball B of the same mass. After the collision, ball A moves at a speed of 1.0 m s⁻¹ at an angle of 45° to the original direction of motion.

(a.i)

State the vertical component of the total momentum of the balls after the collision.

[1]

Markscheme

Zero ✓

(a.ii)

Hence, calculate the vertical component of the velocity of ball B after the collision.

[2]

Markscheme

$m \times 1.0 \times \sin 45 ° + m \times v_{y} = 0$ ✓

$v_{y}$ = «−»0.71 «m s⁻¹» ✓

(b)

Determine the angle θ that the velocity of ball B makes with the initial direction of motion of ball A.

[3]

Markscheme

The use of conservation of momentum in the horizontal direction, e.g. $m \times 1.0 \times \cos 45 ° + m \times v_{x} = m \times 2.0$ ✓

$v_{x} =$ « $2.0 - 1.0 \times \cos 45 ° =$ » $1.3$ «m s⁻¹» ✓

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) = 29 °$ ✓

(c)

Predict whether the collision is elastic.

[4]

Markscheme

Initial kinetic energy $= \frac{1}{2} m {(2.0)}^{2} = 2.0 (m)$

Final kinetic energy $= \frac{1}{2} m ({(1.0)}^{2} + {(0.71)}^{2} + {(1.3)}^{2})$ ✓

$1.6 (m)$ ✓

Final energy is less than the initial energy hence inelastic ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.3ai

(i)

State the vertical component of the total momentum of the balls after the collision.

[1]

Markscheme

Zero ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.3aii

(ii)

Hence, calculate the vertical component of the velocity of ball B after the collision.

[2]

Markscheme

$m \times 1.0 \times \sin 45 ° + m \times v_{y} = 0$ ✓

$v_{y}$ = «−»0.71 «m s⁻¹» ✓

b.

[Maximum mark: 3]

EXE.2.HL.TZ0.3b

(b)

Determine the angle θ that the velocity of ball B makes with the initial direction of motion of ball A.

[3]

Markscheme

The use of conservation of momentum in the horizontal direction, e.g. $m \times 1.0 \times \cos 45 ° + m \times v_{x} = m \times 2.0$ ✓

$v_{x} =$ « $2.0 - 1.0 \times \cos 45 ° =$ » $1.3$ «m s⁻¹» ✓

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) = 29 °$ ✓

c.

[Maximum mark: 4]

EXE.2.HL.TZ0.3c

(c)

Predict whether the collision is elastic.

[4]

Markscheme

Initial kinetic energy $= \frac{1}{2} m {(2.0)}^{2} = 2.0 (m)$

Final kinetic energy $= \frac{1}{2} m ({(1.0)}^{2} + {(0.71)}^{2} + {(1.3)}^{2})$ ✓

$1.6 (m)$ ✓

Final energy is less than the initial energy hence inelastic ✓

4.

[Maximum mark: 6]

EXE.2.HL.TZ0.4

Two curling stones of the same mass collide elastically on a horizontal frictionless surface. Stone A moves with an initial speed v and stone B is initially stationary. The speeds of the stones after the collision are v_A and v_B. Their directions of motion make angles of 70° and 20° with the initial velocity of stone A.

(a.i)

State what is meant by an elastic collision.

[1]

Markscheme

No change in the kinetic energy of the system ✓

(a.ii)

No unbalanced external forces act on the system of the curling stones. Outline why the momentum of the system does not change during the collision.

[1]

Markscheme

From $F = \frac{Δ p}{Δ t}$ , zero net force on the system implies that $Δ p = 0$

OR

From Newton’s third law, the impulse delivered to A is equal but opposite to the impulse delivered to B, hence $Δ p = 0$ for the system ✓

(b)

Show that $v_{B} = \frac{\sin 70 °}{\sin 20 °} v_{A}$ .

[1]

Markscheme

The vertical momentum is zero hence $m v_{A} \sin 70 ° - m v_{B} \sin 20 ° = 0$ ✓

«leading to the expected result» ✓

(c)

Determine v_A. State the answer in terms of v.

[3]

Markscheme

Energy is conserved hence $\frac{1}{2} m {v_{A}}^{2} + \frac{1}{2} m {v_{B}}^{2} = \frac{1}{2} m v^{2}$ ✓

Eliminate $v_{B}$ using the result of part (a), e.g., ${v_{A}}^{2} + {(\frac{\sin 70 °}{\sin 20 °})}^{2} {v_{A}}^{2} = v^{2}$ ✓

$v_{A} = 0.342 v$ ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.4ai

(i)

State what is meant by an elastic collision.

[1]

Markscheme

No change in the kinetic energy of the system ✓

ii.

[Maximum mark: 1]

EXE.2.HL.TZ0.4aii

(ii)

No unbalanced external forces act on the system of the curling stones. Outline why the momentum of the system does not change during the collision.

[1]

Markscheme

From $F = \frac{Δ p}{Δ t}$ , zero net force on the system implies that $Δ p = 0$

OR

From Newton’s third law, the impulse delivered to A is equal but opposite to the impulse delivered to B, hence $Δ p = 0$ for the system ✓

b.

[Maximum mark: 1]

EXE.2.HL.TZ0.4b

(b)

Show that $v_{B} = \frac{\sin 70 °}{\sin 20 °} v_{A}$ .

[1]

Markscheme

The vertical momentum is zero hence $m v_{A} \sin 70 ° - m v_{B} \sin 20 ° = 0$ ✓

«leading to the expected result» ✓

c.

[Maximum mark: 3]

EXE.2.HL.TZ0.4c

(c)

Determine v_A. State the answer in terms of v.

[3]

Markscheme

Energy is conserved hence $\frac{1}{2} m {v_{A}}^{2} + \frac{1}{2} m {v_{B}}^{2} = \frac{1}{2} m v^{2}$ ✓

Eliminate $v_{B}$ using the result of part (a), e.g., ${v_{A}}^{2} + {(\frac{\sin 70 °}{\sin 20 °})}^{2} {v_{A}}^{2} = v^{2}$ ✓

$v_{A} = 0.342 v$ ✓

5.

[Maximum mark: 7]

EXE.2.HL.TZ0.5

A curling stone of mass 17 kg travelling at 2.5 cm s⁻¹ collides with a second stationary curling stone of mass 19 kg. The second curling stone is scattered with a speed of 0.50 cm s⁻¹ at an angle of 30° to the initial direction of the first stone.

(a.i)

Calculate the component of momentum of the first curling stone perpendicular to the initial direction.

[1]

Markscheme

$v_{17} \sin θ = \frac{19 \times 0.5 \times \sin (30)}{17} (= 0.279 m s^{- 1})$ ✓

(a.ii)

Calculate the velocity component of the first curling stone in the initial direction.

[2]

Markscheme

$v_{17} \cos θ = \frac{17 \times 2.5 - 19 \times 0.50 \times \cos (30)}{17}$ ✓

$= 2.02 {m s}^{- 1}$ ✓

(a.iii)

Determine the velocity of the first curling stone.

[2]

Markscheme

2.04 m s⁻¹ ✓

At 7.9° to initial direction ✓

(b)

Deduce whether this collision is elastic.

[2]

Markscheme

Total angle between stones is 38°, angle will be 90° when elastic

OR

Compares kinetic energies in a correct calculation (initial ke = 53 J, final ke = 34 J +2.4 J) ✓

Collision is not elastic ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.5ai

(i)

Calculate the component of momentum of the first curling stone perpendicular to the initial direction.

[1]

Markscheme

$v_{17} \sin θ = \frac{19 \times 0.5 \times \sin (30)}{17} (= 0.279 m s^{- 1})$ ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.5aii

(ii)

Calculate the velocity component of the first curling stone in the initial direction.

[2]

Markscheme

$v_{17} \cos θ = \frac{17 \times 2.5 - 19 \times 0.50 \times \cos (30)}{17}$ ✓

$= 2.02 {m s}^{- 1}$ ✓

iii.

[Maximum mark: 2]

EXE.2.HL.TZ0.5aiii

(iii)

Determine the velocity of the first curling stone.

[2]

Markscheme

2.04 m s⁻¹ ✓

At 7.9° to initial direction ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.5b

(b)

Deduce whether this collision is elastic.

[2]

Markscheme

Total angle between stones is 38°, angle will be 90° when elastic

OR

Compares kinetic energies in a correct calculation (initial ke = 53 J, final ke = 34 J +2.4 J) ✓

Collision is not elastic ✓

6.

[Maximum mark: 5]

EXE.2.HL.TZ0.6

A cannon is used to fire a shell into snow to trigger an avalanche before the snow can cause damage. The mass of the cannon is 1500 kg and the mass of the shell is 15 kg. The shell is projected with an initial speed of 420 m s⁻¹ at an angle of 20° above the horizontal. The cannon is mounted so that it can only recoil horizontally.

(a)

Determine the recoil velocity of the cannon.

[3]

Markscheme

Momentum must be conserved in initial direction of shell (20° above horizontal) ✓

Recoil velocity is 4.2 m s⁻¹ at 20° below horizontal ✓

3.95 m s⁻¹ ✓

(b)

Calculate the initial kinetic energy of the cannon.

[1]

Markscheme

$E_{k} = (\frac{1}{2} m v^{2}) = 11.7 kJ$ ✓

(c)

Suggest what happens to the vertical component of momentum of the cannon when the shell is fired.

[1]

Markscheme

Must be transferred into the ground beneath the cannon OR into the suspension system ✓

a.

[Maximum mark: 3]

EXE.2.HL.TZ0.6a

(a)

Determine the recoil velocity of the cannon.

[3]

Markscheme

Momentum must be conserved in initial direction of shell (20° above horizontal) ✓

Recoil velocity is 4.2 m s⁻¹ at 20° below horizontal ✓

3.95 m s⁻¹ ✓

b.

[Maximum mark: 1]

EXE.2.HL.TZ0.6b

(b)

Calculate the initial kinetic energy of the cannon.

[1]

Markscheme

$E_{k} = (\frac{1}{2} m v^{2}) = 11.7 kJ$ ✓

c.

[Maximum mark: 1]

EXE.2.HL.TZ0.6c

(c)

Suggest what happens to the vertical component of momentum of the cannon when the shell is fired.

[1]

Markscheme

Must be transferred into the ground beneath the cannon OR into the suspension system ✓

7.

[Maximum mark: 6]

EXE.2.HL.TZ0.7

A flywheel of radius $R = 0.15 m$ and mass $M = 1.8 kg$ rotates around the central axis. The moment of inertia of the flywheel is $\frac{1}{2} M R^{2}$ . A thread is wrapped around the flywheel and a time-varying force $F$ is applied to the thread.

The angular velocity of the flywheel increases from 4.0 rad s⁻¹ to 9.0 rad s⁻¹ in a time of 0.24 s.

(a)

Calculate the angular impulse delivered to the flywheel during the acceleration.

[2]

Markscheme

$Δ L = l Δ ω = \frac{1}{2} \times 1.8 \times 0 . 15^{2} (9.0 - 4.0)$ ✓

$0.10$ «N m s» ✓

(b)

Determine the average magnitude of $F$ .

[2]

Markscheme

Average torque $= \frac{Δ L}{Δ t} = \frac{0.10}{0.24} = 0.42$ «N m» ✓

Average force $= \frac{τ}{R} = \frac{0.42}{0.15} = 2.8$ «N» ✓

(c)

State two assumptions of your calculation in part (b).

[2]

Markscheme

No other forces than F provide the torque ✓

The thread unwinds without slipping ✓

The thread is weightless ✓

F is always tangent to the flywheel ✓

a.

[Maximum mark: 2]

EXE.2.HL.TZ0.7a

(a)

Calculate the angular impulse delivered to the flywheel during the acceleration.

[2]

Markscheme

$Δ L = l Δ ω = \frac{1}{2} \times 1.8 \times 0 . 15^{2} (9.0 - 4.0)$ ✓

$0.10$ «N m s» ✓

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.7b

(b)

Determine the average magnitude of $F$ .

[2]

Markscheme

Average torque $= \frac{Δ L}{Δ t} = \frac{0.10}{0.24} = 0.42$ «N m» ✓

Average force $= \frac{τ}{R} = \frac{0.42}{0.15} = 2.8$ «N» ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.7c

(c)

State two assumptions of your calculation in part (b).

[2]

Markscheme

No other forces than F provide the torque ✓

The thread unwinds without slipping ✓

The thread is weightless ✓

F is always tangent to the flywheel ✓

8.

[Maximum mark: 7]

EXE.2.HL.TZ0.8

A ring of mass M = 0.32 kg and radius R = 0.25 m is accelerated from rest by a constant torque of 0.20 N m. The moment of inertia of the ring is MR².

(a)

Calculate:

(a.i)

the angular acceleration of the ring;

[2]

Markscheme

$α = \frac{τ}{l} = \frac{0.20}{0.32 \times 0 . 25^{2}}$ ✓

$10$ «rad s⁻²» ✓

(a.ii)

the angular velocity of the ring after a time of 5.0 s.

[1]

Markscheme

$10 \times 5.0 = 50$ «rad s⁻¹» ✓

(b)

A solid disc of the same mass and radius as the ring is accelerated by the same torque. Compare, without calculation:

(b.i)

the angular impulse delivered to the disc and to the ring during the first 5.0 s.

[2]

Markscheme

The angular impulse is the product of torque and time ✓

Both factors are the same so the angular impulse is the same ✓

(b.ii)

the final kinetic energy of the disc and the ring.

[2]

Markscheme

The disc has a smaller moment of inertia «because its mass is distributed closer to the axis of rotation» ✓

From $E_{k} = \frac{L^{2}}{2 l}$ , the disc will achieve a greater kinetic energy «because $L$ is the same for both» ✓

i.

[Maximum mark: 2]

EXE.2.HL.TZ0.8ai

(i)

the angular acceleration of the ring;

[2]

Markscheme

$α = \frac{τ}{l} = \frac{0.20}{0.32 \times 0 . 25^{2}}$ ✓

$10$ «rad s⁻²» ✓

ii.

[Maximum mark: 1]

EXE.2.HL.TZ0.8aii

(ii)

the angular velocity of the ring after a time of 5.0 s.

[1]

Markscheme

$10 \times 5.0 = 50$ «rad s⁻¹» ✓

i.

[Maximum mark: 2]

EXE.2.HL.TZ0.8bi

(i)

the angular impulse delivered to the disc and to the ring during the first 5.0 s.

[2]

Markscheme

The angular impulse is the product of torque and time ✓

Both factors are the same so the angular impulse is the same ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.8bii

(ii)

the final kinetic energy of the disc and the ring.

[2]

Markscheme

The disc has a smaller moment of inertia «because its mass is distributed closer to the axis of rotation» ✓

From $E_{k} = \frac{L^{2}}{2 l}$ , the disc will achieve a greater kinetic energy «because $L$ is the same for both» ✓

9.

[Maximum mark: 7]

EXE.2.HL.TZ0.9

The propellor of an model plane is driven by an electric motor that is mounted inside the plane. This propellor is modelled as a rod rotating about an axis through the centre of the rod at right angles to the length of the rod. The moment of inertia for such a rod is $\frac{1}{12} M L^{2}$ where $M$ is the mass of the rod and $L$ is the total length of the rod.

(a)

For the propellor, $L = 5.0 cm$ and $M = 0.035 kg$ .

Calculate the moment of inertia of the propellor.

[1]

Markscheme

$7.3 \times 10^{- 6} {kg m}^{2}$ ✓

(b)

The propellor is at rest when the electric motor is switched on. The net average torque acting on the propellor due to the motor and resistive forces is $3.5 \times 10^{- 2} kg m^{2} s^{- 2}$ . The final speed of the propellor is 190 revolutions per second.

(b.i)

Calculate the angular impulse that acts on the propellor.

[2]

Markscheme

190 rev s⁻¹ = 1190 rad s⁻¹✓

$Δ L = Δ (I ω) = 7.3 \times 10^{- 6} \times 1190 = 8.7 \times 10^{- 3} {kg m}^{2} s^{- 1}$ ✓

(b.ii)

Calculate, using your answer to (b)(i), the time taken by the propellor to attain this rotational speed.

[2]

Markscheme

$Δ t = \frac{Δ L}{τ}$ used ✓

0.25 s ✓

(b.iii)

State and explain the effect of the angular impulse on the body of the aeroplane.

[2]

Markscheme

As the motor is internal, angular momentum is conserved (ignoring the torques due to resistive forces) ✓

The body of the plane will (try to) rotate in the opposite direction to the propellor ✓

a.

[Maximum mark: 1]

EXE.2.HL.TZ0.9a

(a)

For the propellor, $L = 5.0 cm$ and $M = 0.035 kg$ .

Calculate the moment of inertia of the propellor.

[1]

Markscheme

$7.3 \times 10^{- 6} {kg m}^{2}$ ✓

i.

[Maximum mark: 2]

EXE.2.HL.TZ0.9bi

(i)

Calculate the angular impulse that acts on the propellor.

[2]

Markscheme

190 rev s⁻¹ = 1190 rad s⁻¹✓

$Δ L = Δ (I ω) = 7.3 \times 10^{- 6} \times 1190 = 8.7 \times 10^{- 3} {kg m}^{2} s^{- 1}$ ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.9bii

(ii)

Calculate, using your answer to (b)(i), the time taken by the propellor to attain this rotational speed.

[2]

Markscheme

$Δ t = \frac{Δ L}{τ}$ used ✓

0.25 s ✓

iii.

[Maximum mark: 2]

EXE.2.HL.TZ0.9biii

(iii)

State and explain the effect of the angular impulse on the body of the aeroplane.

[2]

Markscheme

As the motor is internal, angular momentum is conserved (ignoring the torques due to resistive forces) ✓

The body of the plane will (try to) rotate in the opposite direction to the propellor ✓

a.

[Maximum mark: 3]

EXE.2.HL.TZ0.a

Determine the fractional number of throws for which the three most likely macrostates occur.

[3]

Markscheme

Recognises that 4 heads and 6 tails also required ✓

Total number of microstates = 672 ✓

Fractional number = 672/1024 = 0.66 ✓

Allow ecf for MP3

b.

[Maximum mark: 2]

EXE.2.HL.TZ0.b

Determine the wavelength of the incident photon.

[2]

Markscheme

Energy of incident photon =« $15.9 \times 10^{3} + 500 =$ » $16.4 \times 10^{3}$ «eV» ✓

Wavelength of incident photon =« $\frac{1.24 \times 10^{- 6}}{16.4 \times 10^{3}} =$ » $7.56 \times 10^{- 11}$ «m» ✓

c.

[Maximum mark: 2]

EXE.2.HL.TZ0.c

Outline why the results of the experiment are inconsistent with the wave model of electromagnetic radiation.

[2]

Markscheme

The wavelength of the X-rays changes ✓

According to the wave model, the wavelength of the incident and scattered X-rays should be the same ✓

d.

[Maximum mark: 4]

EXE.2.HL.TZ0.d

Another explosion is produced at X. The sound from this explosion is detected twice at the geophone at Z. Some sound travels directly from X to Z through clay along path 2. Other sound travels through clay via Y along path 3.

The vertical thickness of the clay layer is d. The distance XZ is 80.0 m.

The time between the arrival of the sounds due to the path difference is 6.67 ms.

Calculate d.

[4]

Markscheme

distance difference $= (3000 \times 0.00667 =) 19.8$ m ✓

½ distance difference $= 9.9$ m so YZ $= 49.9$ m ✓

$d = (\sqrt{{YZ}^{2} - {(\frac{XZ}{2})}^{2}}) = \sqrt{49 . 9^{2} - 40^{2}}$ ✓

29.8 m ✓

OR

Recognises situation as (almost) 3:4:5 triangle ✓

30 m (1 sf answer only accepted in this route) ✓

i.

[Maximum mark: 1]

EXE.2.HL.TZ0.i

orbital speed;

[1]

Markscheme

« $\sqrt{\frac{6.67 \times 10^{- 11} \times 5.97 \times 10^{24}}{6.70 \times 10^{6}}} =$ » $7.71 \times 10^{3}$ «m s⁻¹» ✓

ii.

[Maximum mark: 2]

EXE.2.HL.TZ0.ii

Outline why the wavelength of the photon has changed.

[2]

Markscheme

(Energy of photon inversely prop to wavelength)

photon transfers some of its energy to the electron. ✓

If its energy decreases so its wavelength increases. ✓

iii.

[Maximum mark: 2]

EXE.2.HL.TZ0.iii

Deduce the scattering angle for the photon.

[2]

Markscheme

(for this interaction)

$Δ λ = \frac{h}{m_{e} c} (1 - \cos θ) = \frac{h}{m_{e} c}$ and therefore $(1 - \cos θ)$ must equal 1 ✓

so cos theta = 0 and theta = 90 deg ✓

iv.

[Maximum mark: 2]

EXE.2.HL.TZ0.iv

Determine, in days, the half-life of ${}_{52}^{132}{Te}$ .

[2]

Markscheme

Slope =«−» $2.5 \times 10^{- 6}$ «s⁻¹» ✓

$T_{\frac{1}{2}} =$ « $\frac{\ln 2}{2.5 \times 10^{- 6} \times 60 \times 60 \times 24} =$ » 3.2 «days» ✓

10.

[Maximum mark: 7]

EXE.2.SL.TZ0.10

A sealed container of volume 0.35 m³ holds 1.6 mol of a monoatomic gas that can be treated as ideal. The temperature of the gas is 320 K.

One mole of the gas has a mass of 4.0 × 10⁻² kg.

(a)

Calculate the pressure of the gas in the container.

[1]

Markscheme

$p = \frac{n R T}{V}$ seen $= 12 kPa$ ✓

(b)

Determine the mass of the gas in the container.

[1]

Markscheme

$6.4 \times 10^{- 2} kg$ ✓

(c)

Calculate the average translational speed of the gas particles.

[1]

Markscheme

$v = \sqrt{\frac{3 P V}{M}} = 450 {m s}^{- 1}$ ✓

(d)

The temperature of the gas in the container is increased.

Explain, using the kinetic theory, how this change leads to a change in pressure in the container.

[4]

Markscheme

increased temperature means increased average KE and hence increased average translational speed ✓

This increases the momentum transfer at the walls for each collision / mv is greater per collision ✓

This increases the frequency of collisions at the walls / particles cover the distance between walls more quickly ✓

Ideas that $F_{wall} = \frac{Δ p}{Δ t}$ AND $P = \frac{F}{A}$ (can be in words) so that force increases and pressure increases ✓

a.

[Maximum mark: 1]

EXE.2.SL.TZ0.10a

(a)

Calculate the pressure of the gas in the container.

[1]

Markscheme

$p = \frac{n R T}{V}$ seen $= 12 kPa$ ✓

b.

[Maximum mark: 1]

EXE.2.SL.TZ0.10b

(b)

Determine the mass of the gas in the container.

[1]

Markscheme

$6.4 \times 10^{- 2} kg$ ✓

c.

[Maximum mark: 1]

EXE.2.SL.TZ0.10c

(c)

Calculate the average translational speed of the gas particles.

[1]

Markscheme

$v = \sqrt{\frac{3 P V}{M}} = 450 {m s}^{- 1}$ ✓

d.

[Maximum mark: 4]

EXE.2.SL.TZ0.10d

(d)

The temperature of the gas in the container is increased.

Explain, using the kinetic theory, how this change leads to a change in pressure in the container.

[4]

Markscheme

increased temperature means increased average KE and hence increased average translational speed ✓

This increases the momentum transfer at the walls for each collision / mv is greater per collision ✓

This increases the frequency of collisions at the walls / particles cover the distance between walls more quickly ✓

Ideas that $F_{wall} = \frac{Δ p}{Δ t}$ AND $P = \frac{F}{A}$ (can be in words) so that force increases and pressure increases ✓

11.

[Maximum mark: 5]

EXE.2.SL.TZ0.11

(a)

A comet orbits the Sun in an elliptical orbit. A and B are two positions of the comet.

Explain, with reference to Kepler’s second law of planetary motion, the change in the kinetic energy of the comet as it moves from A to B.

[3]

Markscheme

The areas swept out in unit time by the Sun-comet line are the same at A and B ✓

At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓

A decrease in speed means that the kinetic energy also decreases ✓

(b)

An asteroid (minor planet) orbits the Sun in a circular orbit of radius 4.5 × 10⁸ km. The radius of Earth’s orbit is 1.5 × 10⁸ km. Calculate, in years, the orbital period of the asteroid.

[2]

Markscheme

An attempt to use Kepler’s 3^rd law, e.g., ${(\frac{T}{1})}^{2} = {(\frac{4.5}{1.5})}^{3}$ ✓

$T =$ « ${(\frac{4.5}{1.5})}^{1.5} =$ » 5.2 «years» ✓

a.

[Maximum mark: 3]

EXE.2.SL.TZ0.11a

(a)

A comet orbits the Sun in an elliptical orbit. A and B are two positions of the comet.

Explain, with reference to Kepler’s second law of planetary motion, the change in the kinetic energy of the comet as it moves from A to B.

[3]

Markscheme

The areas swept out in unit time by the Sun-comet line are the same at A and B ✓

At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓

A decrease in speed means that the kinetic energy also decreases ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.11b

(b)

An asteroid (minor planet) orbits the Sun in a circular orbit of radius 4.5 × 10⁸ km. The radius of Earth’s orbit is 1.5 × 10⁸ km. Calculate, in years, the orbital period of the asteroid.

[2]

Markscheme

An attempt to use Kepler’s 3^rd law, e.g., ${(\frac{T}{1})}^{2} = {(\frac{4.5}{1.5})}^{3}$ ✓

$T =$ « ${(\frac{4.5}{1.5})}^{1.5} =$ » 5.2 «years» ✓

12.

[Maximum mark: 7]

EXE.2.SL.TZ0.12

One of Kepler’s laws suggests that for moons that have circular orbits around a planet:

$\frac{T^{2}}{4 π r^{3}} = k$

where $T$ is the orbital period of the moon, $r$ is the radius of its circular orbit about the planet, and $k$ is a constant.

(a)

Show that $k = \frac{1}{G M}$ .

[2]

Markscheme

Equates centripetal force (with Newton’s law of gravitation $m r ω^{2} = \frac{G M m}{r^{2}}$ )

OR

$T = \frac{2 π}{ω}$ ✓

Uses both equation correctly with clear re-arrangement ✓

(b)

The table gives data relating to the two moons of Mars.

Moon	T / hour	r / Mm
Phobos	7.66	9.38
Deimos	30.4	-

Determine r for Deimos.

[2]

Markscheme

$r_{De}^{3} = \frac{T_{De}^{2} r_{Ph}^{3}}{T_{Ph}^{2}}$ seen or correct substitution ✓

23.5 Mm ✓

(c)

Determine the mass of Mars.

[3]

Markscheme

Converts T to 27.6 ks and converts to m from Mm ✓

$k = 7.33 \times 10^{- 14}$ «s² m⁻³» ✓

« $M = \frac{1}{k G}$ » $= 2.04 \times 10^{23}$ «kg» ✓

MP1 can be implicit

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.12a

(a)

Show that $k = \frac{1}{G M}$ .

[2]

Markscheme

Equates centripetal force (with Newton’s law of gravitation $m r ω^{2} = \frac{G M m}{r^{2}}$ )

OR

$T = \frac{2 π}{ω}$ ✓

Uses both equation correctly with clear re-arrangement ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.12b

(b)

The table gives data relating to the two moons of Mars.

Moon	T / hour	r / Mm
Phobos	7.66	9.38
Deimos	30.4	-

Determine r for Deimos.

[2]

Markscheme

$r_{De}^{3} = \frac{T_{De}^{2} r_{Ph}^{3}}{T_{Ph}^{2}}$ seen or correct substitution ✓

23.5 Mm ✓

c.

[Maximum mark: 3]

EXE.2.SL.TZ0.12c

(c)

Determine the mass of Mars.

[3]

Markscheme

Converts T to 27.6 ks and converts to m from Mm ✓

$k = 7.33 \times 10^{- 14}$ «s² m⁻³» ✓

« $M = \frac{1}{k G}$ » $= 2.04 \times 10^{23}$ «kg» ✓

MP1 can be implicit

13.

[Maximum mark: 6]

EXE.2.SL.TZ0.13

(a)

Show that $T \propto r^{\frac{3}{2}}$ for the planets in a solar system where $T$ is the orbital period of a planet and $r$ is the radius of circular orbit of planet about its sun.

[2]

Markscheme

Equates centripetal force (with Newton’s law gravitation $m r ω^{2} = \frac{G M m}{r^{2}}$ )

AND

$T = \frac{2 π}{ω}$ ✓

leads to $T^{2} = r^{3} \times (\frac{4 π^{2}}{G M})$ hence result ✓

(b.i)

Outline what is meant by one astronomical unit (1 AU)

[1]

Markscheme

«mean» Distance from centre of Sun to centre of Earth ✓

OR

Suitable ratio in terms of parsec and arcsecond ✓

(b.ii)

Pluto is a dwarf planet of the Sun that orbits at a distance of 5.9 × 10⁹ km from the Sun. Determine, in years, the orbital period of Pluto.

[3]

Markscheme

$T_{Pluto}^{2} = T_{Earth}^{2} \frac{r_{Pluto}^{3}}{r_{Earth}^{3}}$ used ✓

Earth orbital radius = 1.5 × 10¹¹ m (from AU) AND uses 1 earth year (in any units) ✓

247 years ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.13a

(a)

Show that $T \propto r^{\frac{3}{2}}$ for the planets in a solar system where $T$ is the orbital period of a planet and $r$ is the radius of circular orbit of planet about its sun.

[2]

Markscheme

Equates centripetal force (with Newton’s law gravitation $m r ω^{2} = \frac{G M m}{r^{2}}$ )

AND

$T = \frac{2 π}{ω}$ ✓

leads to $T^{2} = r^{3} \times (\frac{4 π^{2}}{G M})$ hence result ✓

i.

[Maximum mark: 1]

EXE.2.SL.TZ0.13bi

(i)

Outline what is meant by one astronomical unit (1 AU)

[1]

Markscheme

«mean» Distance from centre of Sun to centre of Earth ✓

OR

Suitable ratio in terms of parsec and arcsecond ✓

ii.

[Maximum mark: 3]

EXE.2.SL.TZ0.13bii

(ii)

Pluto is a dwarf planet of the Sun that orbits at a distance of 5.9 × 10⁹ km from the Sun. Determine, in years, the orbital period of Pluto.

[3]

Markscheme

$T_{Pluto}^{2} = T_{Earth}^{2} \frac{r_{Pluto}^{3}}{r_{Earth}^{3}}$ used ✓

Earth orbital radius = 1.5 × 10¹¹ m (from AU) AND uses 1 earth year (in any units) ✓

247 years ✓

14.

[Maximum mark: 7]

EXE.2.SL.TZ0.14

Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.

The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10⁻⁵ N.

(a)

State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.

[2]

Markscheme

From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓

$2.0 \times 10^{- 5} N$ ✓

(b)

Calculate the current in wire Q.

[2]

Markscheme

$\frac{2.0 \times 10^{- 5}}{0.50} = \frac{4 π \times 10^{- 7} \times 5.0 \times I_{Q}}{2 π \times 0.10}$ ✓

$I_{Q} = 4.0$ «A» ✓

(c)

Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.

The net magnetic force on wire Q is now zero.

(c.i)

State the direction of the current in R, relative to the current in P.

[1]

Markscheme

Opposite ✓

(c.ii)

Deduce the current in R.

[2]

Markscheme

The force on Q due to R must have the same magnitude «but opposite direction» as the force on Q due to P ✓

The distance is halved therefore one half of the current is needed to produce the same force, so 2.5 A ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.14a

(a)

State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.

[2]

Markscheme

From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓

$2.0 \times 10^{- 5} N$ ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.14b

(b)

Calculate the current in wire Q.

[2]

Markscheme

$\frac{2.0 \times 10^{- 5}}{0.50} = \frac{4 π \times 10^{- 7} \times 5.0 \times I_{Q}}{2 π \times 0.10}$ ✓

$I_{Q} = 4.0$ «A» ✓

i.

[Maximum mark: 1]

EXE.2.SL.TZ0.14ci

(i)

State the direction of the current in R, relative to the current in P.

[1]

Markscheme

Opposite ✓

ii.

[Maximum mark: 2]

EXE.2.SL.TZ0.14cii

(ii)

Deduce the current in R.

[2]

Markscheme

The force on Q due to R must have the same magnitude «but opposite direction» as the force on Q due to P ✓

The distance is halved therefore one half of the current is needed to produce the same force, so 2.5 A ✓

15.

[Maximum mark: 6]

EXE.2.SL.TZ0.15

(a)

State the fundamental SI units for permeability of free space, $μ_{0}$ .

[1]

Markscheme

kg m s⁻² A⁻² ✓

A long straight wire carries a current of 2.0 A. A square conducting loop ABCD of side length 0.20 m is placed near the straight wire, with side AB at a distance of 0.30 m from the wire. There is a current of 1.0 A in the loop. The directions of the currents are shown.

(b)

State the direction, due to the current in the straight wire, of the

(b.i)

magnetic field at A;

[1]

Markscheme

Into the page ✓

(b.ii)

magnetic force on section AB of the loop.

[1]

Markscheme

Repulsive / to the right ✓

(c)

Determine the

(c.i)

magnitude of the net force acting on the loop;

[2]

Markscheme

$(\frac{4 π \times 10^{- 7} \times 2.0 \times 1.0}{2 π \times 0.30} - \frac{4 π \times 10^{- 7} \times 2.0 \times 1.0}{2 π \times 0.50}) \times 0.20$ ✓

$1.1 \times 10^{- 7}$ «N» ✓

(c.ii)

direction of the net force acting on the loop.

[1]

Markscheme

Repulsive / to the right ✓

a.

[Maximum mark: 1]

EXE.2.SL.TZ0.15a

(a)

State the fundamental SI units for permeability of free space, $μ_{0}$ .

[1]

Markscheme

kg m s⁻² A⁻² ✓

i.

[Maximum mark: 1]

EXE.2.SL.TZ0.15bi

(i)

magnetic field at A;

[1]

Markscheme

Into the page ✓

ii.

[Maximum mark: 1]

EXE.2.SL.TZ0.15bii

(ii)

magnetic force on section AB of the loop.

[1]

Markscheme

Repulsive / to the right ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.15ci

(i)

magnitude of the net force acting on the loop;

[2]

Markscheme

$(\frac{4 π \times 10^{- 7} \times 2.0 \times 1.0}{2 π \times 0.30} - \frac{4 π \times 10^{- 7} \times 2.0 \times 1.0}{2 π \times 0.50}) \times 0.20$ ✓

$1.1 \times 10^{- 7}$ «N» ✓

ii.

[Maximum mark: 1]

EXE.2.SL.TZ0.15cii

(ii)

direction of the net force acting on the loop.

[1]

Markscheme

Repulsive / to the right ✓

16.

[Maximum mark: 7]

EXE.2.SL.TZ0.16

Two parallel conducting wires both of length 0.25 m are arranged 0.20 m apart in a circuit. The resistance of one wire is 15 Ω and the resistance of the other wire is 30 Ω. The current in the 15 Ω wire is 10 A.

Assume that the magnetic force due to the Earth can be ignored.

(a)

Determine the magnetic force acting on the 15 Ω wire due to the current in the 30 Ω wire.

[4]

Markscheme

Use of combination of resistors OR $V = I R$ ✓

To show that current in 30 Ω wire is 5.0 A ✓

$\frac{F}{I} = \frac{2 \times 10^{- 7} I_{15} I_{30}}{r} = \frac{2 \times 10^{- 7} \times 10 \times 5}{0.20}$ ✓

$(F = 5 \times 10^{- 5} \times 0.25 =) 1.3 \times 10^{- 5}$ N ✓

(b)

The magnetic field strength of Earth’s field at the location of the wires is 45 μT.

Discuss the assumption made in this question.

[3]

Markscheme

Use of $F = B I l$ ✓

$45 \times 10^{- 6} \times 10 \times 0.25 = 1.15 \times 10^{- 4} N$ ✓

Concludes that the assumption is not valid ✓

a.

[Maximum mark: 4]

EXE.2.SL.TZ0.16a

(a)

Determine the magnetic force acting on the 15 Ω wire due to the current in the 30 Ω wire.

[4]

Markscheme

Use of combination of resistors OR $V = I R$ ✓

To show that current in 30 Ω wire is 5.0 A ✓

$\frac{F}{I} = \frac{2 \times 10^{- 7} I_{15} I_{30}}{r} = \frac{2 \times 10^{- 7} \times 10 \times 5}{0.20}$ ✓

$(F = 5 \times 10^{- 5} \times 0.25 =) 1.3 \times 10^{- 5}$ N ✓

b.

[Maximum mark: 3]

EXE.2.SL.TZ0.16b

(b)

The magnetic field strength of Earth’s field at the location of the wires is 45 μT.

Discuss the assumption made in this question.

[3]

Markscheme

Use of $F = B I l$ ✓

$45 \times 10^{- 6} \times 10 \times 0.25 = 1.15 \times 10^{- 4} N$ ✓

Concludes that the assumption is not valid ✓

17.

[Maximum mark: 7]

EXE.2.SL.TZ0.17

Two parallel wires A and B both carry an electrical current into the page.

(a.i)

Draw the magnetic field lines due to A.

[2]

Markscheme

At least one circle centred on centre of wire A
AND
indication of clockwise direction ✓

More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓

(a.ii)

State and explain, using your diagram, why a force acts on B due to A in the plane of the paper.

[3]

Markscheme

B lies in magnetic field of A OWTTE ✓

Explained use of appropriate rule together with drawn indication of rule operating in this case ✓

To show that force on B is to left and in plane of paper ✓
OR
Magnetic field lines of B merge with those of A to give combined field line pattern ✓

Sketch of combined pattern to show null point somewhere on line between wires. ✓

Wires will move to reduce stored energy and this is achieved by moving together so force on B is to left ✓

(b)

Both wires are 7.5 m long and are 0.25 m apart. The current in both wires is 12 A. Determine the force that acts on one wire due to the other.

[2]

Markscheme

$\frac{F}{l} = \frac{2 \times 10^{- 7} I_{A} I_{B}}{r} = \frac{2 \times 10^{- 7} \times 12^{2}}{0.25}$ ✓

$8.6 \times 10^{- 4} N$ ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.17ai

(i)

Draw the magnetic field lines due to A.

[2]

Markscheme

At least one circle centred on centre of wire A
AND
indication of clockwise direction ✓

More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓

ii.

[Maximum mark: 3]

EXE.2.SL.TZ0.17aii

(ii)

State and explain, using your diagram, why a force acts on B due to A in the plane of the paper.

[3]

Markscheme

B lies in magnetic field of A OWTTE ✓

Explained use of appropriate rule together with drawn indication of rule operating in this case ✓

To show that force on B is to left and in plane of paper ✓
OR
Magnetic field lines of B merge with those of A to give combined field line pattern ✓

Sketch of combined pattern to show null point somewhere on line between wires. ✓

Wires will move to reduce stored energy and this is achieved by moving together so force on B is to left ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.17b

(b)

Both wires are 7.5 m long and are 0.25 m apart. The current in both wires is 12 A. Determine the force that acts on one wire due to the other.

[2]

Markscheme

$\frac{F}{l} = \frac{2 \times 10^{- 7} I_{A} I_{B}}{r} = \frac{2 \times 10^{- 7} \times 12^{2}}{0.25}$ ✓

$8.6 \times 10^{- 4} N$ ✓

18.

[Maximum mark: 8]

EXE.2.SL.TZ0.18

One possible fission reaction of uranium-235 $({}_{92}^{235}U)$ is

${}_{92}^{235}U + n \to {}_{55}^{137}{Cs} + {}_{37}^{95}{Rb} + 4 n$

(a)

Outline, with reference to the decay equation above, the role of chain reactions in the operation of a nuclear power station.

[3]

Markscheme

The four neutrons released in the reaction may initiate further fissions ✓

«If sufficient U-235 is available,» the reaction is self-sustained ✓

Allowing for the continuous production of energy ✓

The number of neutrons available is controlled with control rods «to maintain the desired reaction rate» ✓

The following data are given:

Binding energy per nucleon of ${}_{92}^{235}U$ = 7.591 MeV

Binding energy per nucleon of caesium-137 $({}_{55}^{137}{Cs})$ = 8.389 MeV

Binding energy per nucleon of rubidium-95 $({}_{37}^{95}{Rb})$ = 8.460 MeV

(b)

Calculate, in MeV, the energy released in the reaction.

[2]

Markscheme

$137 \times 8.389 + 95 \times 8.460 - 235 \times 7.591$ ✓

169 «MeV» ✓

(c)

Two nuclides present in spent nuclear fuel ${}_{55}^{137}{Cs}$ are and cerium-144 ( ${}_{58}^{144}{Ce}$ ). The initial activity of a sample of pure ${}_{58}^{144}{Ce}$ is about 40 times greater than the activity of the same amount of pure ${}_{55}^{137}{Cs}$ .

Discuss which of the two nuclides is more likely to require long-term storage once removed from the reactor.

[3]

Markscheme

«For the same number of nuclei,» the activity is inversely related to half-life ✓

Thus ${}_{55}^{137}{Cs}$ has a longer half-life and will likely require longer storage ✓

Half-lives of their decay products need also be considered when planning storage ✓

a.

[Maximum mark: 3]

EXE.2.SL.TZ0.18a

(a)

Outline, with reference to the decay equation above, the role of chain reactions in the operation of a nuclear power station.

[3]

Markscheme

The four neutrons released in the reaction may initiate further fissions ✓

«If sufficient U-235 is available,» the reaction is self-sustained ✓

Allowing for the continuous production of energy ✓

The number of neutrons available is controlled with control rods «to maintain the desired reaction rate» ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.18b

The following data are given:

Binding energy per nucleon of ${}_{92}^{235}U$ = 7.591 MeV

Binding energy per nucleon of caesium-137 $({}_{55}^{137}{Cs})$ = 8.389 MeV

Binding energy per nucleon of rubidium-95 $({}_{37}^{95}{Rb})$ = 8.460 MeV

(b)

Calculate, in MeV, the energy released in the reaction.

[2]

Markscheme

$137 \times 8.389 + 95 \times 8.460 - 235 \times 7.591$ ✓

169 «MeV» ✓

c.

[Maximum mark: 3]

EXE.2.SL.TZ0.18c

(c)

Two nuclides present in spent nuclear fuel ${}_{55}^{137}{Cs}$ are and cerium-144 ( ${}_{58}^{144}{Ce}$ ). The initial activity of a sample of pure ${}_{58}^{144}{Ce}$ is about 40 times greater than the activity of the same amount of pure ${}_{55}^{137}{Cs}$ .

Discuss which of the two nuclides is more likely to require long-term storage once removed from the reactor.

[3]

Markscheme

«For the same number of nuclei,» the activity is inversely related to half-life ✓

Thus ${}_{55}^{137}{Cs}$ has a longer half-life and will likely require longer storage ✓

Half-lives of their decay products need also be considered when planning storage ✓

19.

[Maximum mark: 7]

EXE.2.SL.TZ0.19

(a)

Compare and contrast spontaneous and neutron-induced nuclear fission.

[2]

Markscheme

Spontaneous fission occurs with no external influence, neutron-induced fission requires an interaction with a neutron «of appropriate energy» ✓

Both result in the release of energy
OR
both have a large number of possible pairs of products ✓

(b)

Every neutron-induced fission reaction of uranium-235 releases an energy of about 200 MeV. A nuclear power station transfers an energy of about 2.4 GJ per second.

Determine the mass of uranium-235 that undergoes fission in one day in this power station.

[3]

Markscheme

Fissions per day $= \frac{2.4 \times 10^{9} \times 60 \times 60 \times 24}{200 \times 10^{6} \times 1.6 \times 10^{- 19}}$ « $= 6.5 \times 10^{24}$ » ✓

Mass of uranium $= \frac{6.5 \times 10^{24}}{6.05 \times 10^{23}} \times 0.235$ ✓

2.5 «kg» ✓

(c)

State two properties of the products of nuclear fission due to which the spent nuclear fuel needs to be kept safe.

[2]

Markscheme

Have relatively short half-lives / high activity ✓

Their decay products are «usually» also radioactive ✓

Volatile / chemically active ✓

Biologically active / easily absorbed by living matter ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.19a

(a)

Compare and contrast spontaneous and neutron-induced nuclear fission.

[2]

Markscheme

Spontaneous fission occurs with no external influence, neutron-induced fission requires an interaction with a neutron «of appropriate energy» ✓

Both result in the release of energy
OR
both have a large number of possible pairs of products ✓

b.

[Maximum mark: 3]

EXE.2.SL.TZ0.19b

(b)

Every neutron-induced fission reaction of uranium-235 releases an energy of about 200 MeV. A nuclear power station transfers an energy of about 2.4 GJ per second.

Determine the mass of uranium-235 that undergoes fission in one day in this power station.

[3]

Markscheme

Fissions per day $= \frac{2.4 \times 10^{9} \times 60 \times 60 \times 24}{200 \times 10^{6} \times 1.6 \times 10^{- 19}}$ « $= 6.5 \times 10^{24}$ » ✓

Mass of uranium $= \frac{6.5 \times 10^{24}}{6.05 \times 10^{23}} \times 0.235$ ✓

2.5 «kg» ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.19c

(c)

State two properties of the products of nuclear fission due to which the spent nuclear fuel needs to be kept safe.

[2]

Markscheme

Have relatively short half-lives / high activity ✓

Their decay products are «usually» also radioactive ✓

Volatile / chemically active ✓

Biologically active / easily absorbed by living matter ✓

2.

[Maximum mark: 20]

EXE.2.SL.TZ0.2

An Alpine village uses an electric tram system to transport visitors from a lower station up to an upper station at the village. The length of the tramline is 3.0 km and the gradient of the tramline is a constant 10°.

The tram has a weight of 5.0 × 10⁴ N and can carry a maximum of 75 passengers of average weight 710 N.

The energy is supplied to each tram through a single overhead cable with a resistance per unit length of 0.024 Ω km⁻¹. The tram rails are used for the return path of the current. The return path and the connections from the cable to the electric motor in the tram have negligible resistance.

The power supply maintains a constant emf of 500 V between the rails and the cable at the upper station.

Assume that the current through the motor is constant at 600 A and that the motor efficiency is always 0.90 for the entire range of voltages available to the tram.

(a)

A tram is just leaving the lower railway station.

Determine, as the train leaves the lower station,

(a.i)

the pd across the motor of the tram,

[2]

Markscheme

Resistance of cable = 0.072 Ω ✓

Pd is (500 − 0.072 × 600) = 457 V ✓

(a.ii)

the mechanical power output of the motor.

[2]

Markscheme

Power input = 457 × 600 = 274 kW ✓

Power output = 0.9 × 274 = 247 kW ✓

(b)

Discuss the variation in the power output of the motor with distance from the lower station.

[2]

Markscheme

The pd across the motor increases as the tram travels up the track ✓

(As the current is constant), the power output also rises ✓

(c)

The total friction in the system acting on the tram is equivalent to an opposing force of 750 N.

For one particular journey, the tram is full of passengers.

Estimate the maximum speed v of the tram as it leaves the lower station.

[4]

Markscheme

Total weight of tram = 75 × 710 + 5 × 10⁴ = 1.03 × 10⁵ N ✓

Total force down track = 750 + 1.03 × 10⁵ sin (10) = 1.87 × 10⁴ N ✓

Use of P= F × v ✓

(v = 247 000 ÷ 1.87 × 10⁴)= 13 m s⁻¹ ✓

(d)

The tram travels at v throughout the journey. Two trams are available so that one is returning to the lower station on another line while the other is travelling to the village. The journeys take the same time.

It takes 1.5 minutes to unload and 1.5 minutes to load each tram. Ignore the time taken to accelerate the tram at the beginning and end of the journey.

Estimate the maximum number of passengers that can be carried up to the village in one hour.

[4]

Markscheme

Time for run = s/v = 3000 ÷ 13.2 = 227 s ✓

3 minutes loading = 180 s

So one trip = 407 s ✓

And there are 3600/407 trips per hour = 8.84 ✓

So 8 complete trips with 75 = 600 passengers ✓

(e)

There are eight wheels on each tram with a brake system for each wheel. A pair of brake pads clamp firmly onto an annulus made of steel.

The train comes to rest from speed v. Ignore the energy transferred to the brake pads and the change in the gravitational potential energy of the tram during the braking.

Calculate the temperature change in each steel annulus as the tram comes to rest.

Data for this question

The inner radius of the annulus is 0.40 m and the outer radius is 0.50 m.

The thickness of the annulus is 25 mm.

The density of the steel is 7860 kg m⁻³

The specific heat capacity of the steel is 420 J kg⁻¹ K⁻¹

[4]

Markscheme

Work leading to volume = 7.1 x 10⁻³ m³ ✓

Work leading to mass of steel = 55 .8 kg ✓

Kinetic Energy transferred per annulus = $\frac{\frac{1}{2} m v^{2}}{8} = \frac{1}{16} \times \frac{1.03 \times 10^{5}}{9.81} \times 13^{2}$

= 110 kJ ✓

$Δ T = \frac{E_{k}}{m c} = \frac{1.1 \times 10^{5}}{55.8 \times 420} = 4.7$ K ✓

(f)

The speed of the tram is measured by detecting a beam of microwaves of wavelength 2.8 cm reflected from the rear of the tram as it moves away from the station. Predict the change in wavelength of the microwaves at the stationary microwave detector in the station.

[2]

Markscheme

Use of $\frac{Δ λ}{λ} \approx \frac{v}{c}$ ✓

1.2 nm ✓

20.

[Maximum mark: 5]

EXE.2.SL.TZ0.20

(a)

State one source of the radioactive waste products from nuclear fission reactions.

[1]

Markscheme

Fission fragments from the fuel rods

OR activated materials in (e.g.) fuel rod casings

OR nuclei formed by neutron activation from U-235

OR stated products, e.g. Pu, U-236 etc. ✓

(b)

Outline how this waste is treated after it has been removed from the fission reactor.

[4]

Markscheme

Waste (fuel rod) is placed in cooling ponds for a number of years ✓

After most active products have decayed the uranium is separated to be recycled/reprocessed ✓

The remaining highly active waste is vitrified / made into a solid form ✓

And stored (deep) underground ✓

a.

[Maximum mark: 1]

EXE.2.SL.TZ0.20a

(a)

State one source of the radioactive waste products from nuclear fission reactions.

[1]

Markscheme

Fission fragments from the fuel rods

OR activated materials in (e.g.) fuel rod casings

OR nuclei formed by neutron activation from U-235

OR stated products, e.g. Pu, U-236 etc. ✓

b.

[Maximum mark: 4]

EXE.2.SL.TZ0.20b

(b)

Outline how this waste is treated after it has been removed from the fission reactor.

[4]

Markscheme

Waste (fuel rod) is placed in cooling ponds for a number of years ✓

After most active products have decayed the uranium is separated to be recycled/reprocessed ✓

The remaining highly active waste is vitrified / made into a solid form ✓

And stored (deep) underground ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.2ai

(i)

the pd across the motor of the tram,

[2]

Markscheme

Resistance of cable = 0.072 Ω ✓

Pd is (500 − 0.072 × 600) = 457 V ✓

ii.

[Maximum mark: 2]

EXE.2.SL.TZ0.2aii

(ii)

the mechanical power output of the motor.

[2]

Markscheme

Power input = 457 × 600 = 274 kW ✓

Power output = 0.9 × 274 = 247 kW ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.2b

(b)

Discuss the variation in the power output of the motor with distance from the lower station.

[2]

Markscheme

The pd across the motor increases as the tram travels up the track ✓

(As the current is constant), the power output also rises ✓

c.

[Maximum mark: 4]

EXE.2.SL.TZ0.2c

(c)

The total friction in the system acting on the tram is equivalent to an opposing force of 750 N.

For one particular journey, the tram is full of passengers.

Estimate the maximum speed v of the tram as it leaves the lower station.

[4]

Markscheme

Total weight of tram = 75 × 710 + 5 × 10⁴ = 1.03 × 10⁵ N ✓

Total force down track = 750 + 1.03 × 10⁵ sin (10) = 1.87 × 10⁴ N ✓

Use of P= F × v ✓

(v = 247 000 ÷ 1.87 × 10⁴)= 13 m s⁻¹ ✓

d.

[Maximum mark: 4]

EXE.2.SL.TZ0.2d

(d)

The tram travels at v throughout the journey. Two trams are available so that one is returning to the lower station on another line while the other is travelling to the village. The journeys take the same time.

It takes 1.5 minutes to unload and 1.5 minutes to load each tram. Ignore the time taken to accelerate the tram at the beginning and end of the journey.

Estimate the maximum number of passengers that can be carried up to the village in one hour.

[4]

Markscheme

Time for run = s/v = 3000 ÷ 13.2 = 227 s ✓

3 minutes loading = 180 s

So one trip = 407 s ✓

And there are 3600/407 trips per hour = 8.84 ✓

So 8 complete trips with 75 = 600 passengers ✓

e.

[Maximum mark: 4]

EXE.2.SL.TZ0.2e

(e)

There are eight wheels on each tram with a brake system for each wheel. A pair of brake pads clamp firmly onto an annulus made of steel.

The train comes to rest from speed v. Ignore the energy transferred to the brake pads and the change in the gravitational potential energy of the tram during the braking.

Calculate the temperature change in each steel annulus as the tram comes to rest.

Data for this question

The inner radius of the annulus is 0.40 m and the outer radius is 0.50 m.

The thickness of the annulus is 25 mm.

The density of the steel is 7860 kg m⁻³

The specific heat capacity of the steel is 420 J kg⁻¹ K⁻¹

[4]

Markscheme

Work leading to volume = 7.1 x 10⁻³ m³ ✓

Work leading to mass of steel = 55 .8 kg ✓

Kinetic Energy transferred per annulus = $\frac{\frac{1}{2} m v^{2}}{8} = \frac{1}{16} \times \frac{1.03 \times 10^{5}}{9.81} \times 13^{2}$

= 110 kJ ✓

$Δ T = \frac{E_{k}}{m c} = \frac{1.1 \times 10^{5}}{55.8 \times 420} = 4.7$ K ✓

f.

[Maximum mark: 2]

EXE.2.SL.TZ0.2f

(f)

The speed of the tram is measured by detecting a beam of microwaves of wavelength 2.8 cm reflected from the rear of the tram as it moves away from the station. Predict the change in wavelength of the microwaves at the stationary microwave detector in the station.

[2]

Markscheme

Use of $\frac{Δ λ}{λ} \approx \frac{v}{c}$ ✓

1.2 nm ✓

3.

[Maximum mark: 8]

EXE.2.SL.TZ0.3

A steel pot containing water is placed on an electric hot plate that is preheated to a temperature of 180 °C. The initial temperature of the water in the pot is 10 °C.

The base of the pot has a surface area of 0.15 m² and a thickness of 5.0 mm. The coefficient of thermal conductivity of the material of the pot is 45 W m⁻¹ K⁻¹.

(a)

Calculate:

(a.i)

the initial temperature gradient through the base of the pot. State an appropriate unit for your answer.

[2]

Markscheme

$\frac{180 - 10}{0.005} = 3.4 \times 10^{4}$ ✓

K m⁻¹ ✓

(a.ii)

the initial rate, in kW, of thermal energy transfer by conduction through the base of the pot.

[1]

Markscheme

45 × 0.15 × 3.4 × 10⁴ = 230 «kW» ✓

(b)

The electrical power rating of the hot plate is 1 kW. Comment, with reference to this value, on your answer in (a)(ii).

[3]

Markscheme

The answer is unrealistically large / impossible to sustain ✓

Due to much lower actual power, a lower temperature gradient through the base of the pot is quickly established ✓

The surface of the hot plate becomes colder from contact with pot
OR
there is a temperature gradient also through the hot plate ✓

(c)

Describe how thermal energy is distributed throughout the volume of the water in the pot.

[2]

Markscheme

By means of convection currents ✓

That arise due to density difference between hot and cold water ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.3ai

(i)

the initial temperature gradient through the base of the pot. State an appropriate unit for your answer.

[2]

Markscheme

$\frac{180 - 10}{0.005} = 3.4 \times 10^{4}$ ✓

K m⁻¹ ✓

ii.

[Maximum mark: 1]

EXE.2.SL.TZ0.3aii

(ii)

the initial rate, in kW, of thermal energy transfer by conduction through the base of the pot.

[1]

Markscheme

45 × 0.15 × 3.4 × 10⁴ = 230 «kW» ✓

b.

[Maximum mark: 3]

EXE.2.SL.TZ0.3b

(b)

The electrical power rating of the hot plate is 1 kW. Comment, with reference to this value, on your answer in (a)(ii).

[3]

Markscheme

The answer is unrealistically large / impossible to sustain ✓

Due to much lower actual power, a lower temperature gradient through the base of the pot is quickly established ✓

The surface of the hot plate becomes colder from contact with pot
OR
there is a temperature gradient also through the hot plate ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.3c

(c)

Describe how thermal energy is distributed throughout the volume of the water in the pot.

[2]

Markscheme

By means of convection currents ✓

That arise due to density difference between hot and cold water ✓

4.

[Maximum mark: 6]

EXE.2.SL.TZ0.4

A sealed bottle contains 0.50 kg of water at an initial temperature of 60 °C. The bottle is made of glass of thickness 3.0 mm and thermal conductivity 0.90 W m⁻¹ K⁻¹.

(a)

The temperature of the air outside of the bottle is 20 °C. The surface area of the bottle is 4.0 × 10⁻² m². Calculate the initial rate of thermal energy transfer by conduction through the bottle.

[2]

Markscheme

$0.90 \times 4.0 \times 10^{- 2} \times \frac{60 - 20}{0.003}$ ✓

480 «W» ✓

(b)

Explain why the rate calculated in part (a) is decreasing.

[2]

Markscheme

The temperature gradient decreases as the water cools down ✓

The rate of energy transfer is proportional to the temperature gradient ✓

(c)

Estimate the initial rate of the change of the temperature of the water in the bottle. State your answer in K s⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

[2]

Markscheme

$480 = 4200 \times 0.50 \times \frac{Δ T}{Δ t}$ ✓

$\frac{Δ T}{Δ t} = 0.23$ «K s⁻¹» ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.4a

(a)

The temperature of the air outside of the bottle is 20 °C. The surface area of the bottle is 4.0 × 10⁻² m². Calculate the initial rate of thermal energy transfer by conduction through the bottle.

[2]

Markscheme

$0.90 \times 4.0 \times 10^{- 2} \times \frac{60 - 20}{0.003}$ ✓

480 «W» ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.4b

(b)

Explain why the rate calculated in part (a) is decreasing.

[2]

Markscheme

The temperature gradient decreases as the water cools down ✓

The rate of energy transfer is proportional to the temperature gradient ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.4c

(c)

Estimate the initial rate of the change of the temperature of the water in the bottle. State your answer in K s⁻¹. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

[2]

Markscheme

$480 = 4200 \times 0.50 \times \frac{Δ T}{Δ t}$ ✓

$\frac{Δ T}{Δ t} = 0.23$ «K s⁻¹» ✓

5.

[Maximum mark: 6]

EXE.2.SL.TZ0.5

The ends of a vertical column of water are maintained at different temperatures T_t and T_b both above the freezing point.

Energy transfer by radiation in this arrangement is negligible.

(a)

Discuss the mechanism that accounts for the greatest rate of energy transfer when:

(a.i)

T_t > T_b

[2]

Markscheme

Conduction identified ✓

energy transfer through interaction of particles in liquid at atomic scale ✓

(a.ii)

T_b > T_t

[2]

Markscheme

Convection identified ✓

energy transfer through movement of bodies of liquid at different densities ✓

(b)

The liquid now freezes so that the vertical column is entirely of ice. Suggest how your answer to (a)(ii) will change.

[2]

Markscheme

the solid cannot now move relative to material above it ✓

so conduction only ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.5ai

(i)

T_t > T_b

[2]

Markscheme

Conduction identified ✓

energy transfer through interaction of particles in liquid at atomic scale ✓

ii.

[Maximum mark: 2]

EXE.2.SL.TZ0.5aii

(ii)

T_b > T_t

[2]

Markscheme

Convection identified ✓

energy transfer through movement of bodies of liquid at different densities ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.5b

(b)

The liquid now freezes so that the vertical column is entirely of ice. Suggest how your answer to (a)(ii) will change.

[2]

Markscheme

the solid cannot now move relative to material above it ✓

so conduction only ✓

6.

[Maximum mark: 8]

EXE.2.SL.TZ0.6

A rod is formed from two metal rods XY and YZ of identical dimensions. End X and end Z are at different temperatures.

(a)

The side of the rod can be unlagged or ideally lagged. Explain the difference in energy transfer for these two cases.

[3]

Markscheme

When ideally lagged, no energy transfer can occur through the sides of the bar. ✓

All the power input/ energy input per second at one end will emerge at the other end. ✓

When unlagged, energy transfer occurs from the sides of the bar and the power /energy input per second at input > the energy output per second at the other end. ✓

Max 1 if answer does not refer to rate of energy transfer in MP2 and MP3.

(b)

Rod XYZ is ideally lagged. The thermal conductivity of XY is k and the thermal conductivity of YZ is 2k. End X is at 90 °C and end Z is at 45 °C

(b.i)

Calculate the temperature at Y.

[3]

Markscheme

idea that $\frac{d Q}{d t}$ is same in both bars because lagged ✓

work to show that $90 - θ = 2 (θ - 45)$
OR
Temperature difference across XY is twice temperature difference across YZ ✓

solves to show that $θ = 60 ° C$ ✓

(b.ii)

The temperatures are now reversed so that X is at 45 °C and Z is at 90 °C. Show that the rate of energy transfer is unchanged.

[2]

Markscheme

repeats calculation to show that θ = 75 °C ✓

temperature difference across YZ is still 15 K which gives the same rate of energy transfer (but in opposite direction) ✓

a.

[Maximum mark: 3]

EXE.2.SL.TZ0.6a

(a)

The side of the rod can be unlagged or ideally lagged. Explain the difference in energy transfer for these two cases.

[3]

Markscheme

When ideally lagged, no energy transfer can occur through the sides of the bar. ✓

All the power input/ energy input per second at one end will emerge at the other end. ✓

When unlagged, energy transfer occurs from the sides of the bar and the power /energy input per second at input > the energy output per second at the other end. ✓

Max 1 if answer does not refer to rate of energy transfer in MP2 and MP3.

i.

[Maximum mark: 3]

EXE.2.SL.TZ0.6bi

(i)

Calculate the temperature at Y.

[3]

Markscheme

idea that $\frac{d Q}{d t}$ is same in both bars because lagged ✓

work to show that $90 - θ = 2 (θ - 45)$
OR
Temperature difference across XY is twice temperature difference across YZ ✓

solves to show that $θ = 60 ° C$ ✓

ii.

[Maximum mark: 2]

EXE.2.SL.TZ0.6bii

(ii)

The temperatures are now reversed so that X is at 45 °C and Z is at 90 °C. Show that the rate of energy transfer is unchanged.

[2]

Markscheme

repeats calculation to show that θ = 75 °C ✓

temperature difference across YZ is still 15 K which gives the same rate of energy transfer (but in opposite direction) ✓

7.

[Maximum mark: 7]

EXE.2.SL.TZ0.7

(a)

State two assumptions of the kinetic model of an ideal gas that refer to intermolecular collisions.

[2]

Markscheme

The collisions are elastic ✓

The time for a collision is much shorter than the time between collisions ✓

The intermolecular forces are only present during collisions ✓

(b)

Discuss how the motion of the molecules of a gas gives rise to pressure in the gas.

[3]

Markscheme

The momentum of a molecule changes when it collides with a container wall ✓

From $F = \frac{Δ p}{Δ t}$ and Newton’s third law, this leads to a force exerted on the wall by the molecule ✓

The average force exerted by all the molecules on a unit area of the wall is equivalent to pressure ✓

(c)

The average speed of the molecules of a gas is 500 m s⁻¹. The density of the gas is 1.2 kg m⁻³. Calculate, in kPa, the pressure of the gas.

[2]

Markscheme

$\frac{1}{3} \times 1.2 \times 500^{2}$ ✓

100 «kPa» ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.7a

(a)

State two assumptions of the kinetic model of an ideal gas that refer to intermolecular collisions.

[2]

Markscheme

The collisions are elastic ✓

The time for a collision is much shorter than the time between collisions ✓

The intermolecular forces are only present during collisions ✓

b.

[Maximum mark: 3]

EXE.2.SL.TZ0.7b

(b)

Discuss how the motion of the molecules of a gas gives rise to pressure in the gas.

[3]

Markscheme

The momentum of a molecule changes when it collides with a container wall ✓

From $F = \frac{Δ p}{Δ t}$ and Newton’s third law, this leads to a force exerted on the wall by the molecule ✓

The average force exerted by all the molecules on a unit area of the wall is equivalent to pressure ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.7c

(c)

The average speed of the molecules of a gas is 500 m s⁻¹. The density of the gas is 1.2 kg m⁻³. Calculate, in kPa, the pressure of the gas.

[2]

Markscheme

$\frac{1}{3} \times 1.2 \times 500^{2}$ ✓

100 «kPa» ✓

8.

[Maximum mark: 7]

EXE.2.SL.TZ0.8

A sample of air in a sealed container has a pressure of 1.8 × 10⁵ Pa and a density of 2.0 kg m⁻³.

(a)

Calculate the average translational speed of air molecules.

[2]

Markscheme

$1.8 \times 10^{5} = \frac{1}{3} \times 2.0 \times v^{2}$ ✓

$v = 520$ «m s⁻¹» ✓

(b)

The air is a mixture of nitrogen, oxygen and other gases. Explain why the component gases of air in the container have different average translational speeds.

[3]

Markscheme

Average kinetic energy of the molecules is determined by the temperature only ✓

The mass of a molecule is different for each component gas ✓

From $E_{k} = \frac{1}{2} m v^{2}$ , the same $E_{k}$ and different mass implies a different average velocity ✓

(c)

The temperature of the sample is increased without a change in pressure. Outline the effect it has on the density of the gas.

[2]

Markscheme

ALTERNATIVE 1

The average translational speed increases «because T increases» ✓

From $P = \frac{1}{3} ρ v^{2}$ , the density decreases «to keep $P$ constant» ✓

ALTERNATIVE 2

From the ideal gas law, the volume of the gas increases ✓

Since $ρ = \frac{m}{v}$ and $m$ is constant, the density decreases ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.8a

(a)

Calculate the average translational speed of air molecules.

[2]

Markscheme

$1.8 \times 10^{5} = \frac{1}{3} \times 2.0 \times v^{2}$ ✓

$v = 520$ «m s⁻¹» ✓

b.

[Maximum mark: 3]

EXE.2.SL.TZ0.8b

(b)

The air is a mixture of nitrogen, oxygen and other gases. Explain why the component gases of air in the container have different average translational speeds.

[3]

Markscheme

Average kinetic energy of the molecules is determined by the temperature only ✓

The mass of a molecule is different for each component gas ✓

From $E_{k} = \frac{1}{2} m v^{2}$ , the same $E_{k}$ and different mass implies a different average velocity ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.8c

(c)

The temperature of the sample is increased without a change in pressure. Outline the effect it has on the density of the gas.

[2]

Markscheme

ALTERNATIVE 1

The average translational speed increases «because T increases» ✓

From $P = \frac{1}{3} ρ v^{2}$ , the density decreases «to keep $P$ constant» ✓

ALTERNATIVE 2

From the ideal gas law, the volume of the gas increases ✓

Since $ρ = \frac{m}{v}$ and $m$ is constant, the density decreases ✓

9.

[Maximum mark: 8]

EXE.2.SL.TZ0.9

(a)

Outline how the concept of absolute zero of temperature is interpreted in terms of:

(a.i)

the ideal gas law,

[1]

Markscheme

it is the temperature at which the volume

OR

the pressure extrapolates to zero (can be shown by sketch) ✓

(a.ii)

the kinetic energy of particles in an ideal gas.

[1]

Markscheme

it is the temperature at which all the (random) motion stops

OR

at which all the motion can be extrapolated to stop

OR

at which the kinetic energy of all particles is zero ✓

(b)

A container holds a mixture of argon and helium atoms at a temperature of 37 °C.

Calculate the average translational speed of the argon atoms.

The molar mass of argon is 4.0 × 10⁻² kg mol⁻¹.

[4]

Markscheme

Use of $\frac{4.0 \times 10^{- 2}}{N_{A}} (= 6.6 \times 10^{- 26} kg)$ ✓

Work showing that $v (= \sqrt{\frac{3 p}{ρ} = \sqrt{\frac{3 p V}{m}}}) = \sqrt{\frac{3 k T}{m}}$ ✓

Correct substitution AND conversion to K (310 K) ✓

430/440 «m s⁻¹» ✓

(c)

Discuss how the mean kinetic energy of the argon atoms in the mixture compares with that of the helium atoms.

[2]

Markscheme

the gases are in the same container at the same temperature so are in equilibrium ✓

they must have the same mean/average kinetic energy ✓

i.

[Maximum mark: 1]

EXE.2.SL.TZ0.9ai

(i)

the ideal gas law,

[1]

Markscheme

it is the temperature at which the volume

OR

the pressure extrapolates to zero (can be shown by sketch) ✓

ii.

[Maximum mark: 1]

EXE.2.SL.TZ0.9aii

(ii)

the kinetic energy of particles in an ideal gas.

[1]

Markscheme

it is the temperature at which all the (random) motion stops

OR

at which all the motion can be extrapolated to stop

OR

at which the kinetic energy of all particles is zero ✓

b.

[Maximum mark: 4]

EXE.2.SL.TZ0.9b

(b)

A container holds a mixture of argon and helium atoms at a temperature of 37 °C.

Calculate the average translational speed of the argon atoms.

The molar mass of argon is 4.0 × 10⁻² kg mol⁻¹.

[4]

Markscheme

Use of $\frac{4.0 \times 10^{- 2}}{N_{A}} (= 6.6 \times 10^{- 26} kg)$ ✓

Work showing that $v (= \sqrt{\frac{3 p}{ρ} = \sqrt{\frac{3 p V}{m}}}) = \sqrt{\frac{3 k T}{m}}$ ✓

Correct substitution AND conversion to K (310 K) ✓

430/440 «m s⁻¹» ✓

c.

[Maximum mark: 2]

EXE.2.SL.TZ0.9c

(c)

Discuss how the mean kinetic energy of the argon atoms in the mixture compares with that of the helium atoms.

[2]

Markscheme

the gases are in the same container at the same temperature so are in equilibrium ✓

they must have the same mean/average kinetic energy ✓

a.

[Maximum mark: 2]

EXE.2.SL.TZ0.a

Compare and contrast spontaneous and neutron-induced nuclear fission.

[2]

Markscheme

Spontaneous fission occurs with no external influence, neutron-induced fission requires an interaction with a neutron «of appropriate energy» ✓

Both result in the release of energy
OR
both have a large number of possible pairs of products ✓

b.

[Maximum mark: 2]

EXE.2.SL.TZ0.b

An asteroid (minor planet) orbits the Sun in a circular orbit of radius 4.5 × 10⁸ km. The radius of Earth’s orbit is 1.5 × 10⁸ km. Calculate, in years, the orbital period of the asteroid.

[2]

Markscheme

An attempt to use Kepler’s 3^rd law, e.g., ${(\frac{T}{1})}^{2} = {(\frac{4.5}{1.5})}^{3}$ ✓

$T =$ « ${(\frac{4.5}{1.5})}^{1.5} =$ » 5.2 «years» ✓

c.

[Maximum mark: 4]

EXE.2.SL.TZ0.c

The total friction in the system acting on the tram is equivalent to an opposing force of 750 N.

For one particular journey, the tram is full of passengers.

Estimate the maximum speed v of the tram as it leaves the lower station.

[4]

Markscheme

Total weight of tram = 75 × 710 + 5 × 10⁴ = 1.03 × 10⁵ N ✓

Total force down track = 750 + 1.03 × 10⁵ sin (10) = 1.87 × 10⁴ N ✓

Use of P= F × v ✓

(v = 247 000 ÷ 1.87 × 10⁴)= 13 m s⁻¹ ✓

d.

[Maximum mark: 4]

EXE.2.SL.TZ0.d

The tram travels at v throughout the journey. Two trams are available so that one is returning to the lower station on another line while the other is travelling to the village. The journeys take the same time.

It takes 1.5 minutes to unload and 1.5 minutes to load each tram. Ignore the time taken to accelerate the tram at the beginning and end of the journey.

Estimate the maximum number of passengers that can be carried up to the village in one hour.

[4]

Markscheme

Time for run = s/v = 3000 ÷ 13.2 = 227 s ✓

3 minutes loading = 180 s

So one trip = 407 s ✓

And there are 3600/407 trips per hour = 8.84 ✓

So 8 complete trips with 75 = 600 passengers ✓

e.

[Maximum mark: 4]

EXE.2.SL.TZ0.e

There are eight wheels on each tram with a brake system for each wheel. A pair of brake pads clamp firmly onto an annulus made of steel.

The train comes to rest from speed v. Ignore the energy transferred to the brake pads and the change in the gravitational potential energy of the tram during the braking.

Calculate the temperature change in each steel annulus as the tram comes to rest.

Data for this question

The inner radius of the annulus is 0.40 m and the outer radius is 0.50 m.

The thickness of the annulus is 25 mm.

The density of the steel is 7860 kg m⁻³

The specific heat capacity of the steel is 420 J kg⁻¹ K⁻¹

[4]

Markscheme

Work leading to volume = 7.1 x 10⁻³ m³ ✓

Work leading to mass of steel = 55 .8 kg ✓

Kinetic Energy transferred per annulus = $\frac{\frac{1}{2} m v^{2}}{8} = \frac{1}{16} \times \frac{1.03 \times 10^{5}}{9.81} \times 13^{2}$

= 110 kJ ✓

$Δ T = \frac{E_{k}}{m c} = \frac{1.1 \times 10^{5}}{55.8 \times 420} = 4.7$ K ✓

f.

[Maximum mark: 2]

EXE.2.SL.TZ0.f

The speed of the tram is measured by detecting a beam of microwaves of wavelength 2.8 cm reflected from the rear of the tram as it moves away from the station. Predict the change in wavelength of the microwaves at the stationary microwave detector in the station.

[2]

Markscheme

Use of $\frac{Δ λ}{λ} \approx \frac{v}{c}$ ✓

1.2 nm ✓

i.

[Maximum mark: 2]

EXE.2.SL.TZ0.i

the pd across the motor of the tram,

[2]

Markscheme

Resistance of cable = 0.072 Ω ✓

Pd is (500 − 0.072 × 600) = 457 V ✓

ii.

[Maximum mark: 1]

EXE.2.SL.TZ0.ii

the kinetic energy of particles in an ideal gas.

[1]

Markscheme

it is the temperature at which all the (random) motion stops

OR

at which all the motion can be extrapolated to stop

OR

at which the kinetic energy of all particles is zero ✓

1.

[Maximum mark: 1]

SPM.1A.HL.TZ0.1

A car has an initial speed of 16 m s⁻¹. It decelerates at 4.0 m s⁻² until it stops.

What is the distance travelled by the car?

A. 4 m

B. 16 m

C. 32 m

D. 64 m

[1]

Markscheme

C

10.

[Maximum mark: 1]

SPM.1A.HL.TZ0.10

The internal energy of a real gas is

A. zero.

B. equal to the intermolecular potential energy of the particles.

C. equal to the total kinetic energy of the particles.

D. equal to the sum of the intermolecular potential energy and the total kinetic energy of the particles.

[1]

Markscheme

D

11.

[Maximum mark: 1]

SPM.1A.HL.TZ0.11

A gas undergoes one cycle of a cyclic process.

The net change in internal energy of the gas is

A. zero.

B. positive.

C. negative.

D. determined by the initial temperature of the gas.

[1]

Markscheme

A

12.

[Maximum mark: 1]

SPM.1A.HL.TZ0.12

A working refrigerator with the door open is placed in a sealed room.

The entropy of the room

A. is zero.

B. decreases.

C. remains unchanged.

D. increases.

[1]

Markscheme

D

13.

[Maximum mark: 1]

SPM.1A.HL.TZ0.13

The black-body radiation curve of an object at 600 K is shown. The intensity units are arbitrary.

What is the radiation curve of the same object at 450 K?

The original curve is shown with a dashed line.

[1]

Markscheme

A

14.

[Maximum mark: 1]

SPM.1A.HL.TZ0.14

Star X has a luminosity L and an apparent brightness b. Star X is at a distance $d$ from Earth.

Star Y has the same apparent brightness as X but is four times more luminous.

What is the distance of Star Y from Earth?

A. $4 d$

B. $2 d$

C. $\frac{d}{2}$

D. $\frac{d}{4}$

[1]

Markscheme

B

15.

[Maximum mark: 1]

SPM.1A.HL.TZ0.15

Four identical resistors, each of resistance $R$ , are connected as shown.

What is the effective resistance between P and Q?

A. $\frac{3 R}{4}$

B. $R$

C. $\frac{4 R}{3}$

D. $4 R$

[1]

Markscheme

A

16.

[Maximum mark: 1]

SPM.1A.HL.TZ0.16

Conductor X is connected to a cell of emf E. A power of 16 W is dissipated in X.

Conductor Y is made from the same material with the same diameter as X but is twice as long. A cell of emf 2E is connected to Y.

Both cells have negligible internal resistance.

What power is dissipated in Y?

A. 8.0 W

B. 16 W

C. 32 W

D. 64 W

[1]

Markscheme

C

17.

[Maximum mark: 1]

SPM.1A.HL.TZ0.17

Two containers, $X$ and $Y$ , are filled with an ideal gas at the same pressure.

The volume of $X$ is four times the volume of $Y$ . The temperature of $X$ is 327 °C and the temperature of $Y$ is 27 °C.

What is $\frac{amount of substance in X}{amount of substance in Y}$ ?

A. $\frac{1}{8}$

B. $\frac{1}{2}$

C. $2$

D. $8$

[1]

Markscheme

C

18.

[Maximum mark: 1]

SPM.1A.HL.TZ0.18

An electromagnetic wave has a wavelength that is about the size of the diameter of an atom.

What region of the electromagnetic spectrum does the wave belong to?

A. Infrared

B. Visible light

C. Ultraviolet

D. X-ray

[1]

Markscheme

D

19.

[Maximum mark: 1]

SPM.1A.HL.TZ0.19

A particle undergoes simple harmonic motion of period $T$ . At time $t = 0$ the particle is at its equilibrium position.

What is $t$ when the particle is at its greatest distance from the equilibrium position?

A. $\frac{T}{8}$

B. $\frac{T}{2}$

C. $\frac{3 T}{4}$

D. $T$

[1]

Markscheme

C

1.

[Maximum mark: 1]

SPM.1A.SL.TZ0.1

A car has an initial speed of 16 m s⁻¹. It decelerates at 4.0 m s⁻² until it stops.

What is the distance travelled by the car?

A. 4 m

B. 16 m

C. 32 m

D. 64 m

[1]

Markscheme

C

10.

[Maximum mark: 1]

SPM.1A.SL.TZ0.10

Four identical resistors, each of resistance $R$ , are connected as shown.

What is the effective resistance between P and Q?

A. $\frac{3 R}{4}$

B. $R$

C. $\frac{4 R}{3}$

D. $4 R$

[1]

Markscheme

A

11.

[Maximum mark: 1]

SPM.1A.SL.TZ0.11

Conductor X is connected to a cell of emf E. A power of 16 W is dissipated in X.

Conductor Y is made from the same material with the same diameter as X but is twice as long. A cell of emf 2E is connected to Y.

Both cells have negligible internal resistance.

What power is dissipated in Y?

A. 8.0 W

B. 16 W

C. 32 W

D. 64 W

[1]

Markscheme

C

12.

[Maximum mark: 1]

SPM.1A.SL.TZ0.12

Two containers, $X$ and $Y$ , are filled with an ideal gas at the same pressure.

The volume of $X$ is four times the volume of $Y$ . The temperature of $X$ is 327 °C and the temperature of $Y$ is 27 °C.

What is $\frac{amount of substance in X}{amount of substance in Y}$ ?

A. $\frac{1}{8}$

B. $\frac{1}{2}$

C. $2$

D. $8$

[1]

Markscheme

C

13.

[Maximum mark: 1]

SPM.1A.SL.TZ0.13

An electromagnetic wave has a wavelength that is about the size of the diameter of an atom.

What region of the electromagnetic spectrum does the wave belong to?

A. Infrared

B. Visible light

C. Ultraviolet

D. X-ray

[1]

Markscheme

D

14.

[Maximum mark: 1]

SPM.1A.SL.TZ0.14

A particle undergoes simple harmonic motion of period $T$ . At time $t = 0$ the particle is at its equilibrium position.

What is $t$ when the particle is at its greatest distance from the equilibrium position?

A. $\frac{T}{8}$

B. $\frac{T}{2}$

C. $\frac{3 T}{4}$

D. $T$

[1]

Markscheme

C

15.

[Maximum mark: 1]

SPM.1A.SL.TZ0.15

Diagram 1 shows the variation with position of the displacement of a standing wave formed on a string.

Diagram 2 shows the variation with position of the displacement of a travelling wave moving to the right along a string.

Points P, Q, R and S are points on the string.

What is the phase difference between P and Q and the phase difference between R and S?

[1]

Markscheme

C

16.

[Maximum mark: 1]

SPM.1A.SL.TZ0.16

A mass of 0.25 kg hangs from a spring of spring constant 4.0 N m⁻¹.

What is the natural frequency of oscillation for this system?

A. 0.50 Hz

B. 0.64 Hz

C. 1.6 Hz

D. 2.0 Hz

[1]

Markscheme

B

17.

[Maximum mark: 1]

SPM.1A.SL.TZ0.17

Two long parallel wires P and Q are a distance d apart. They each carry a current.

A magnetic force per unit length $F$ acts on P due to Q.

The distance between the wires is increased to 2d and the current in Q is decreased to $\frac{I}{2}$ .

What is the magnetic force per unit length that acts on P due to Q after the changes?

A. $\frac{F}{8}$

B. $\frac{F}{4}$

C. $\frac{F}{2}$

D. $F$

[1]

Markscheme

B

18.

[Maximum mark: 1]

SPM.1A.SL.TZ0.18

Planets X and Y orbit the same star.

The average distance between planet X and the star is five times greater than the average distance between planet Y and the star.

What is $\frac{orbital period of planet X}{orbital period of planet Y}$ ?

A. $\sqrt[3]{5}$

B. $\sqrt{5}$

C. $\sqrt[3]{5^{2}}$

D. $\sqrt{5^{3}}$

[1]

Markscheme

D

19.

[Maximum mark: 1]

SPM.1A.SL.TZ0.19

A charged rod is brought near an initially neutral metal sphere without touching it.

When the sphere is grounded (earthed), there is an electric current for a short time from the sphere to the ground.

The ground connection is then removed.

What are the charge on the rod and the charge induced on the sphere when the connection is removed?

[1]

Markscheme

C

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